[proofplan] We first prove the theorem for compactly supported [absolutely continuous measures](/page/Absolutely%20Continuous%20Measures) that assign zero mass to every affine hyperplane. In that case, after introducing the sphere of oriented affine hyperplane parameters, the signed mass imbalances define a continuous odd map from that sphere to $\mathbb{R}^d$, and the [Borsuk-Ulam theorem](/theorems/6462) gives a zero of this map. For general finite Borel measures, we truncate to large balls, mollify to obtain compactly supported absolutely continuous measures, choose simultaneous bisectors for the approximants, and pass to a convergent subsequence of hyperplanes. Tightness prevents the bisectors from escaping to infinity, and closed half-space inequalities pass to the limit by weak convergence and a small enlargement argument. [/proofplan]

[step:Parametrize oriented affine hyperplanes by the sphere $S^d$] Let \begin{align*} S^d := \{(a,b) \in \mathbb{R}^d \times \mathbb{R} : |a|^2 + b^2 = 1\}. \end{align*} Throughout the proof, let $\mathcal{L}^d$ denote $d$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}^d$ equipped with its Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R}^d)$. For $r>0$, let $B(0,r):=\{x\in\mathbb{R}^d:|x|<r\}$ denote the open Euclidean ball of radius $r$ centered at $0$. For each point $p=(a,b) \in S^d$, define the closed positive and negative regions \begin{align*} E^+(p) := \{x \in \mathbb{R}^d : a \cdot x \geq b\} \end{align*} and \begin{align*} E^-(p) := \{x \in \mathbb{R}^d : a \cdot x \leq b\}. \end{align*} If $a \neq 0$, the common boundary is the affine hyperplane $\{x \in \mathbb{R}^d : a \cdot x = b\}$, equivalently $\{x \in \mathbb{R}^d : x \cdot u = t\}$, where \begin{align*} u := \frac{a}{|a|} \in S^{d-1} \end{align*} and \begin{align*} t := \frac{b}{|a|} \in \mathbb{R}. \end{align*} The antipodal point $-p=(-a,-b)$ reverses the orientation, because \begin{align*} E^+(-p)=E^-(p) \end{align*} and \begin{align*} E^-(-p)=E^+(p). \end{align*} When $a=0$, the two points $(0,1)$ and $(0,-1)$ represent the two hyperplanes at infinity in this compactification: $E^+(0,1)=\varnothing$, $E^-(0,1)=\mathbb{R}^d$, while $E^+(0,-1)=\mathbb{R}^d$ and $E^-(0,-1)=\varnothing$. [/step]

[step:Prove the result for compactly supported absolutely continuous measures]Assume first that each $\mu_i$ is compactly supported, absolutely continuous with respect to $\mathcal{L}^d$, and assigns measure zero to every affine hyperplane. If $\mu_i(\mathbb{R}^d)=0$ for every $i$, then any affine hyperplane satisfies the conclusion, so suppose at least one measure has positive total mass. Define the imbalance map $F:S^d \to \mathbb{R}^d$ by \begin{align*} F(p) := (F_1(p),\dots,F_d(p)). \end{align*} For each $i \in \{1,\dots,d\}$, define the coordinate map $F_i:S^d \to \mathbb{R}$ by \begin{align*} F_i(p) := \mu_i(E^+(p))-\mu_i(E^-(p)) \end{align*} for every $p \in S^d$. The function $F_i$ is continuous. Indeed, if $p_n=(a_n,b_n) \to p=(a,b)$ in $S^d$, then $\mathbb{1}_{E^+(p_n)}(x) \to \mathbb{1}_{E^+(p)}(x)$ and $\mathbb{1}_{E^-(p_n)}(x) \to \mathbb{1}_{E^-(p)}(x)$ for every $x$ not lying on the boundary $\{a \cdot x=b\}$. That boundary has $\mu_i$-measure zero by hypothesis. Since the indicator functions are bounded by $1$ and $\mu_i$ is finite, dominated convergence with respect to the measure $\mu_i$ gives continuity of both $p \mapsto \mu_i(E^+(p))$ and $p \mapsto \mu_i(E^-(p))$. Thus $F$ is continuous. Moreover $F$ is odd, since $E^+(-p)=E^-(p)$ and $E^-(-p)=E^+(p)$, so \begin{align*} F(-p)=-F(p). \end{align*} We use the [Borsuk-Ulam Theorem](/theorems/1170) in the following standard form: every continuous odd map $S^d \to \mathbb{R}^d$ has a zero. Applying this theorem to the continuous odd map $F:S^d \to \mathbb{R}^d$, there exists $p_0=(a_0,b_0) \in S^d$ such that $F(p_0)=0$. If $a_0=0$, then $p_0$ is either $(0,1)$ or $(0,-1)$, and for any index $i$ with $\mu_i(\mathbb{R}^d)>0$ the coordinate $F_i(p_0)$ is either $-\mu_i(\mathbb{R}^d)$ or $\mu_i(\mathbb{R}^d)$, contradicting $F_i(p_0)=0$. Therefore $a_0 \neq 0$. Set $u_0 := a_0/|a_0| \in S^{d-1}$ and $t_0 := b_0/|a_0| \in \mathbb{R}$. For each $i$, \begin{align*} \mu_i(E^+(p_0))=\mu_i(E^-(p_0)). \end{align*} Since the boundary hyperplane has $\mu_i$-measure zero, the two closed half-spaces have union $\mathbb{R}^d$ and overlap of $\mu_i$-measure zero. Hence \begin{align*} \mu_i(E^+(p_0))=\mu_i(E^-(p_0))=\frac{1}{2}\mu_i(\mathbb{R}^d). \end{align*} Thus the hyperplane $\{x \in \mathbb{R}^d : x \cdot u_0=t_0\}$ bisects all the measures in the special case.[/step]

[guided]The compactly supported absolutely continuous case is where the topological argument lives. The point of parametrizing by $S^d$ is that each oriented hyperplane appears together with its opposite orientation as an antipodal point. For $p=(a,b) \in S^d$, define \begin{align*} E^+(p) := \{x \in \mathbb{R}^d : a \cdot x \geq b\} \end{align*} and \begin{align*} E^-(p) := \{x \in \mathbb{R}^d : a \cdot x \leq b\}. \end{align*} If $a \neq 0$, these are the two closed half-spaces bounded by the affine hyperplane $\{x \in \mathbb{R}^d : a \cdot x=b\}$. If $a=0$, the point is one of the two compactification points at infinity. The antipodal point $-p=(-a,-b)$ swaps the two closed regions: \begin{align*} E^+(-p)=E^-(p) \end{align*} and \begin{align*} E^-(-p)=E^+(p). \end{align*} Now define $F:S^d \to \mathbb{R}^d$ by \begin{align*} F(p) := (F_1(p),\dots,F_d(p)) \end{align*} where \begin{align*} F_i(p) := \mu_i(E^+(p))-\mu_i(E^-(p)). \end{align*} The coordinate $F_i(p)$ measures the signed mass imbalance for the $i$th measure. A zero of $F$ means that each measure assigns equal mass to the two closed half-spaces. Because the measures in this step assign zero mass to the separating hyperplane, equality of the two closed half-space masses is the same as exact bisection. We verify continuity using the [dominated convergence theorem](/theorems/4). Let $p_n=(a_n,b_n) \to p=(a,b)$ in $S^d$. For a point $x \in \mathbb{R}^d$ with $a \cdot x \neq b$, the inequalities $a_n \cdot x \geq b_n$ and $a_n \cdot x \leq b_n$ eventually have the same truth values as $a \cdot x \geq b$ and $a \cdot x \leq b$, respectively. Hence \begin{align*} \mathbb{1}_{E^+(p_n)}(x) \to \mathbb{1}_{E^+(p)}(x) \end{align*} and \begin{align*} \mathbb{1}_{E^-(p_n)}(x) \to \mathbb{1}_{E^-(p)}(x) \end{align*} for every $x$ outside the boundary hyperplane $\{x \in \mathbb{R}^d : a \cdot x=b\}$. That boundary has $\mu_i$-measure zero by hypothesis. Since all indicators are bounded by the integrable function $1$ on the finite [measure space](/page/Measure%20Space) $(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d),\mu_i)$, dominated convergence gives \begin{align*} \mu_i(E^+(p_n)) \to \mu_i(E^+(p)) \end{align*} and \begin{align*} \mu_i(E^-(p_n)) \to \mu_i(E^-(p)). \end{align*} Thus each $F_i$ is continuous, and so $F$ is continuous. The oddness is exactly the orientation reversal. Since antipodal points swap the two half-spaces, for every $p \in S^d$, \begin{align*} F_i(-p)=\mu_i(E^+(-p))-\mu_i(E^-(-p))=\mu_i(E^-(p))-\mu_i(E^+(p))=-F_i(p). \end{align*} Therefore $F(-p)=-F(p)$. We now use the [Borsuk-Ulam Theorem](/theorems/1170) in the following standard form: every continuous odd map $S^d \to \mathbb{R}^d$ has a zero. The hypotheses are exactly the two facts just proved: $F:S^d \to \mathbb{R}^d$ is continuous, and $F(-p)=-F(p)$ for every $p \in S^d$. Therefore there is $p_0=(a_0,b_0) \in S^d$ such that $F(p_0)=0$. It remains to check that this zero corresponds to an actual affine hyperplane, not a point at infinity. If $a_0=0$, then $p_0=(0,1)$ or $p_0=(0,-1)$. At $(0,1)$ the positive region is empty and the negative region is all of $\mathbb{R}^d$, so $F_i(0,1)=-\mu_i(\mathbb{R}^d)$. At $(0,-1)$ the signs are reversed, so $F_i(0,-1)=\mu_i(\mathbb{R}^d)$. Since at least one measure has positive total mass, $F(p_0)$ could not be zero. Thus $a_0 \neq 0$. Define $u_0 := a_0/|a_0| \in S^{d-1}$ and $t_0 := b_0/|a_0| \in \mathbb{R}$. The corresponding affine hyperplane is \begin{align*} \{x \in \mathbb{R}^d : x \cdot u_0=t_0\}. \end{align*} For every $i$, the equality $F_i(p_0)=0$ gives \begin{align*} \mu_i(E^+(p_0))=\mu_i(E^-(p_0)). \end{align*} The overlap $E^+(p_0) \cap E^-(p_0)$ is exactly the boundary hyperplane, which has $\mu_i$-measure zero. Since the union is all of $\mathbb{R}^d$, the two equal masses must each be half of the total mass: \begin{align*} \mu_i(E^+(p_0))=\mu_i(E^-(p_0))=\frac{1}{2}\mu_i(\mathbb{R}^d). \end{align*} This proves the compactly supported absolutely continuous case.[/guided]

[step:Approximate the finite Borel measures by compactly supported smooth measures]For each $n \in \mathbb{N}$, choose $R_n>0$ such that, for every $i \in \{1,\dots,d\}$, \begin{align*} \mu_i(\mathbb{R}^d \setminus B(0,R_n)) \leq \frac{1}{n}. \end{align*} This is possible because finite Borel measures on $\mathbb{R}^d$ are tight. Let $\mu_{i,n}$ be the restricted finite Borel measure defined by \begin{align*} \mu_{i,n}(A) := \mu_i(A \cap B(0,R_n)) \end{align*} for every Borel set $A \subset \mathbb{R}^d$. Choose a nonnegative [mollifier](/page/Standard%20Mollifier) $\rho \in C_c^\infty(B(0,1))$ satisfying \begin{align*} \int_{\mathbb{R}^d} \rho(x)\, d\mathcal{L}^d(x)=1. \end{align*} For $\varepsilon_n := 1/n$, define $\rho_{\varepsilon_n}:\mathbb{R}^d \to [0,\infty)$ by \begin{align*} \rho_{\varepsilon_n}(x) := \varepsilon_n^{-d}\rho(x/\varepsilon_n). \end{align*} For each $i$ and $n$, define the smooth density $f_{i,n}:\mathbb{R}^d \to [0,\infty)$ by \begin{align*} f_{i,n}(x) := \int_{\mathbb{R}^d} \rho_{\varepsilon_n}(x-y)\, d\mu_{i,n}(y). \end{align*} Let $\nu_{i,n}$ be the absolutely continuous finite Borel measure defined by \begin{align*} \nu_{i,n}(A) := \int_A f_{i,n}(x)\, d\mathcal{L}^d(x) \end{align*} for every Borel set $A \subset \mathbb{R}^d$. Then $\nu_{i,n}$ is compactly supported in $B(0,R_n+\varepsilon_n)$, is absolutely continuous with respect to $\mathcal{L}^d$, and gives measure zero to every affine hyperplane. Its total mass is \begin{align*} \nu_{i,n}(\mathbb{R}^d)=\mu_{i,n}(\mathbb{R}^d)=\mu_i(B(0,R_n)). \end{align*} Moreover $\nu_{i,n}$ converges [weakly](/page/Weak%20Convergence) to $\mu_i$ as $n \to \infty$. To prove this, fix a bounded continuous map $\varphi:\mathbb{R}^d \to \mathbb{R}$ and define $M := \|\varphi\|_\infty$. If $M=0$, the convergence is immediate, so assume $M>0$. For $y \in \mathbb{R}^d$, define the mollified [test function](/page/Test%20Function) $\varphi_{\varepsilon_n}:\mathbb{R}^d \to \mathbb{R}$ by \begin{align*} \varphi_{\varepsilon_n}(y) := \int_{\mathbb{R}^d} \varphi(y+z)\rho_{\varepsilon_n}(z)\, d\mathcal{L}^d(z). \end{align*} By [Fubini's theorem](/theorems/2961), applied to the Borel function $(x,y) \mapsto \varphi(x)\rho_{\varepsilon_n}(x-y)$ on $(\mathbb{R}^d \times \mathbb{R}^d,\mathcal{B}(\mathbb{R}^d) \otimes \mathcal{B}(\mathbb{R}^d),\mathcal{L}^d \otimes \mu_{i,n})$, the order of integration may be exchanged. Indeed, $\mu_{i,n}$ is supported in $B(0,R_n)$, $\rho_{\varepsilon_n}$ is supported in $B(0,\varepsilon_n)$, and $|\varphi(x)\rho_{\varepsilon_n}(x-y)| \leq M\|\rho_{\varepsilon_n}\|_\infty$ on the finite-measure set $B(0,R_n+\varepsilon_n) \times B(0,R_n)$. Hence \begin{align*} \int_{\mathbb{R}^d} \varphi(x)\, d\nu_{i,n}(x)=\int_{\mathbb{R}^d} \varphi_{\varepsilon_n}(y)\, d\mu_{i,n}(y). \end{align*} Let $\gamma>0$. Since $\mu_i$ is finite on $\mathbb{R}^d$, choose a compact set $K \subset \mathbb{R}^d$ such that \begin{align*} \mu_i(\mathbb{R}^d \setminus K)<\frac{\gamma}{6M}. \end{align*} Let $K+\overline{B}(0,1):=\{y+z:y\in K,\ |z|\leq 1\}$. Because $K+\overline{B}(0,1)$ is compact and $\varphi$ is continuous, $\varphi$ is uniformly continuous on $K+\overline{B}(0,1)$. Hence there exists $N_1 \in \mathbb{N}$ such that, for all $n \geq N_1$ and all $y \in K$, \begin{align*} |\varphi_{\varepsilon_n}(y)-\varphi(y)|<\frac{\gamma}{3(\mu_i(\mathbb{R}^d)+1)}. \end{align*} Also choose $N_2 \in \mathbb{N}$ such that $1/n<\gamma/(6M)$ for all $n \geq N_2$. For $n \geq \max\{N_1,N_2\}$, using $\mu_{i,n}\leq \mu_i$, $|\varphi_{\varepsilon_n}|\leq M$, and $|\varphi|\leq M$, we decompose the error as \begin{align*} \left|\int_{\mathbb{R}^d} \varphi(x)\, d\nu_{i,n}(x)-\int_{\mathbb{R}^d} \varphi(x)\, d\mu_i(x)\right| = \left|\int_{\mathbb{R}^d} \varphi_{\varepsilon_n}(y)\, d\mu_{i,n}(y)-\int_{\mathbb{R}^d} \varphi(y)\, d\mu_i(y)\right|. \end{align*} Since $\mu_i-\mu_{i,n}$ is the restriction of $\mu_i$ to $\mathbb{R}^d\setminus B(0,R_n)$, the triangle inequality gives \begin{align*} \left|\int_{\mathbb{R}^d} \varphi_{\varepsilon_n}\, d\mu_{i,n}-\int_{\mathbb{R}^d} \varphi\, d\mu_i\right| \leq \left|\int_{\mathbb{R}^d}(\varphi_{\varepsilon_n}-\varphi)\,d\mu_{i,n}\right| + \left|\int_{\mathbb{R}^d}\varphi\,d(\mu_{i,n}-\mu_i)\right|. \end{align*} The first term splits over $K$ and its complement, while the second term is only the truncation tail. Thus \begin{align*} \left|\int_{\mathbb{R}^d} \varphi_{\varepsilon_n}(y)\, d\mu_{i,n}(y)-\int_{\mathbb{R}^d} \varphi(y)\, d\mu_i(y)\right| \leq \int_K |\varphi_{\varepsilon_n}(y)-\varphi(y)|\, d\mu_{i,n}(y)+2M\mu_{i,n}(\mathbb{R}^d\setminus K)+M\mu_i(\mathbb{R}^d\setminus B(0,R_n)). \end{align*} The first term is at most $\gamma/3$ by the uniform approximation on $K$. The second is at most $2M\mu_i(\mathbb{R}^d\setminus K)<\gamma/3$, and the third is at most $M/n<\gamma/6$. Hence \begin{align*} \left|\int_{\mathbb{R}^d} \varphi(x)\, d\nu_{i,n}(x)-\int_{\mathbb{R}^d} \varphi(x)\, d\mu_i(x)\right|<\gamma. \end{align*} Therefore \begin{align*} \int_{\mathbb{R}^d} \varphi(x)\, d\nu_{i,n}(x) \to \int_{\mathbb{R}^d} \varphi(x)\, d\mu_i(x). \end{align*}[/step]

[guided]The approximation has two jobs: make the measures smooth enough for the Borsuk-Ulam step, while still converging back to the original finite Borel measures. First choose $R_n>0$ so that every original measure has tail mass at most $1/n$ outside $B(0,R_n)$. This uses tightness of finite Borel measures on $\mathbb{R}^d$. Define the restricted measure $\mu_{i,n}$ by \begin{align*} \mu_{i,n}(A) := \mu_i(A \cap B(0,R_n)) \end{align*} for every Borel set $A \subset \mathbb{R}^d$. Next choose a nonnegative mollifier $\rho \in C_c^\infty(B(0,1))$ with \begin{align*} \int_{\mathbb{R}^d} \rho(x)\, d\mathcal{L}^d(x)=1. \end{align*} For $\varepsilon_n:=1/n$, set \begin{align*} \rho_{\varepsilon_n}(x):=\varepsilon_n^{-d}\rho(x/\varepsilon_n). \end{align*} Define the density $f_{i,n}:\mathbb{R}^d \to [0,\infty)$ by \begin{align*} f_{i,n}(x):=\int_{\mathbb{R}^d}\rho_{\varepsilon_n}(x-y)\,d\mu_{i,n}(y). \end{align*} Then define the Borel measure $\nu_{i,n}$ by \begin{align*} \nu_{i,n}(A):=\int_A f_{i,n}(x)\,d\mathcal{L}^d(x). \end{align*} Because $\rho_{\varepsilon_n}$ is supported in $B(0,\varepsilon_n)$ and $\mu_{i,n}$ is supported in $B(0,R_n)$, the measure $\nu_{i,n}$ is supported in $B(0,R_n+\varepsilon_n)$. Since $\nu_{i,n}$ has density $f_{i,n}$ with respect to $\mathcal{L}^d$, it is absolutely continuous with respect to $\mathcal{L}^d$ and assigns zero mass to every affine hyperplane. Its total mass is preserved by the normalization of the mollifier: \begin{align*} \nu_{i,n}(\mathbb{R}^d)=\mu_{i,n}(\mathbb{R}^d)=\mu_i(B(0,R_n)). \end{align*} It remains to prove weak convergence. Let $\varphi:\mathbb{R}^d\to\mathbb{R}$ be bounded and continuous, and set $M:=\|\varphi\|_\infty$. If $M=0$, the convergence is immediate. Otherwise define \begin{align*} \varphi_{\varepsilon_n}(y):=\int_{\mathbb{R}^d}\varphi(y+z)\rho_{\varepsilon_n}(z)\,d\mathcal{L}^d(z). \end{align*} Fubini's theorem applies because the integrand $(x,y)\mapsto \varphi(x)\rho_{\varepsilon_n}(x-y)$ is bounded by $M\|\rho_{\varepsilon_n}\|_\infty$ on the finite-measure set $B(0,R_n+\varepsilon_n)\times B(0,R_n)$ and vanishes outside it. Hence \begin{align*} \int_{\mathbb{R}^d}\varphi(x)\,d\nu_{i,n}(x)=\int_{\mathbb{R}^d}\varphi_{\varepsilon_n}(y)\,d\mu_{i,n}(y). \end{align*} Fix $\gamma>0$. Choose a compact set $K\subset\mathbb{R}^d$ with \begin{align*} \mu_i(\mathbb{R}^d\setminus K)<\frac{\gamma}{6M}. \end{align*} Let $K+\overline{B}(0,1):=\{y+z:y\in K,\ |z|\leq 1\}$. The set $K+\overline{B}(0,1)$ is compact, so $\varphi$ is uniformly continuous there. Therefore, for all sufficiently large $n$ and all $y\in K$, \begin{align*} |\varphi_{\varepsilon_n}(y)-\varphi(y)|<\frac{\gamma}{3(\mu_i(\mathbb{R}^d)+1)}. \end{align*} Also take $n$ large enough that $1/n<\gamma/(6M)$. Since $\mu_i-\mu_{i,n}$ is the restriction of $\mu_i$ to $\mathbb{R}^d\setminus B(0,R_n)$, first write \begin{align*} \left|\int_{\mathbb{R}^d}\varphi_{\varepsilon_n}\,d\mu_{i,n}-\int_{\mathbb{R}^d}\varphi\,d\mu_i\right| \leq \left|\int_{\mathbb{R}^d}(\varphi_{\varepsilon_n}-\varphi)\,d\mu_{i,n}\right| + \left|\int_{\mathbb{R}^d}\varphi\,d(\mu_{i,n}-\mu_i)\right|. \end{align*} Now split the first integral over $K$ and $\mathbb{R}^d\setminus K$. Using $\mu_{i,n}\leq\mu_i$, $|\varphi_{\varepsilon_n}|\leq M$, and $|\varphi|\leq M$, we obtain \begin{align*} \left|\int_{\mathbb{R}^d}\varphi(x)\,d\nu_{i,n}(x)-\int_{\mathbb{R}^d}\varphi(x)\,d\mu_i(x)\right|\leq \int_K |\varphi_{\varepsilon_n}(y)-\varphi(y)|\,d\mu_{i,n}(y)+2M\mu_{i,n}(\mathbb{R}^d\setminus K)+M\mu_i(\mathbb{R}^d\setminus B(0,R_n)). \end{align*} The three terms are bounded by $\gamma/3$, $\gamma/3$, and $\gamma/6$, respectively. Thus the absolute difference is less than $\gamma$ for all sufficiently large $n$. Since $\gamma>0$ was arbitrary, \begin{align*} \int_{\mathbb{R}^d}\varphi(x)\,d\nu_{i,n}(x)\to\int_{\mathbb{R}^d}\varphi(x)\,d\mu_i(x). \end{align*} This proves $\nu_{i,n}\rightharpoonup\mu_i$ weakly.[/guided]

[step:Choose bisectors for the approximating measures and extract a bounded subsequence]By the special case already proved, for each $n$ there are $u_n \in S^{d-1}$ and $t_n \in \mathbb{R}$ such that, for every $i$, \begin{align*} \nu_{i,n}(\{x \in \mathbb{R}^d : x \cdot u_n \geq t_n\})=\frac{1}{2}\nu_{i,n}(\mathbb{R}^d) \end{align*} and \begin{align*} \nu_{i,n}(\{x \in \mathbb{R}^d : x \cdot u_n \leq t_n\})=\frac{1}{2}\nu_{i,n}(\mathbb{R}^d). \end{align*} If $\mu_i(\mathbb{R}^d)=0$ for all $i$, any affine hyperplane proves the theorem. Otherwise choose an index $i_0$ such that $\mu_{i_0}(\mathbb{R}^d)>0$. We claim that the sequence $(t_n)$ is bounded. If not, pass to a subsequence with $|t_n| \to \infty$. Since $(\nu_{i_0,n})$ converges weakly to the finite measure $\mu_{i_0}$ on the locally compact Polish space $\mathbb{R}^d$, weak convergence implies eventual tightness. Equivalently, for every $\eta>0$ there exists $R>0$ such that \begin{align*} \nu_{i_0,n}(\mathbb{R}^d \setminus B(0,R))<\eta \end{align*} for all sufficiently large $n$. Taking $\eta := \mu_{i_0}(\mathbb{R}^d)/4$ and then $n$ large enough that $|t_n|>R$, one of the two bisected half-spaces contains only points outside $B(0,R)$: if $t_n>R$, this is $\{x:x\cdot u_n \geq t_n\}$; if $t_n<-R$, this is $\{x:x\cdot u_n \leq t_n\}$. Hence that half-space has $\nu_{i_0,n}$-mass less than $\eta$. But its mass equals $\nu_{i_0,n}(\mathbb{R}^d)/2$, and $\nu_{i_0,n}(\mathbb{R}^d) \to \mu_{i_0}(\mathbb{R}^d)$, a contradiction for large $n$. Thus $(t_n)$ is bounded. Since $S^{d-1}$ is compact, after passing to a subsequence there exist $u \in S^{d-1}$ and $t \in \mathbb{R}$ such that \begin{align*} u_n \to u \end{align*} and \begin{align*} t_n \to t. \end{align*}[/step]

[guided]The approximating measures have exact bisectors, but those bisectors could a priori move farther and farther away from the origin. We prevent this by using one measure with positive total mass. Choose $i_0 \in \{1,\dots,d\}$ such that $\mu_{i_0}(\mathbb{R}^d)>0$. Because $\nu_{i_0,n}$ converges weakly to $\mu_{i_0}$ on $\mathbb{R}^d$, the sequence $(\nu_{i_0,n})_{n \in \mathbb{N}}$ is eventually tight. Indeed, fix $\eta>0$. Since $\mu_{i_0}$ is finite, choose $R>0$ such that $\mu_{i_0}(\mathbb{R}^d \setminus B(0,R))<\eta/2$. Applying the [Portmanteau Theorem](/theorems/1171) to the [closed set](/page/Closed%20Set) $\mathbb{R}^d \setminus B(0,R)$ gives \begin{align*} \limsup_{n \to \infty} \nu_{i_0,n}(\mathbb{R}^d \setminus B(0,R)) \leq \mu_{i_0}(\mathbb{R}^d \setminus B(0,R))<\frac{\eta}{2}. \end{align*} Thus $\nu_{i_0,n}(\mathbb{R}^d \setminus B(0,R))<\eta$ for all sufficiently large $n$. This says that the approximating mass cannot escape to infinity. Set \begin{align*} \eta := \frac{1}{4}\mu_{i_0}(\mathbb{R}^d). \end{align*} If $|t_n| \to \infty$ along a subsequence, then for all sufficiently large $n$ we have $|t_n|>R$. If $t_n>R$, every point satisfying $x \cdot u_n \geq t_n$ lies outside $B(0,R)$, since $|x \cdot u_n| \leq |x|$ and $|u_n|=1$. If $t_n<-R$, every point satisfying $x \cdot u_n \leq t_n$ lies outside $B(0,R)$ for the same reason. Thus one of the two bisected half-spaces has $\nu_{i_0,n}$-mass less than $\eta$. But the bisection property gives that the same half-space has mass \begin{align*} \frac{1}{2}\nu_{i_0,n}(\mathbb{R}^d). \end{align*} Since $\nu_{i_0,n}(\mathbb{R}^d) \to \mu_{i_0}(\mathbb{R}^d)$, this is larger than $\eta$ for all sufficiently large $n$, a contradiction. Therefore $(t_n)$ is bounded. The sequence $(u_n)$ lies in the compact sphere $S^{d-1}$, and $(t_n)$ lies in a bounded subset of $\mathbb{R}$. By [sequential compactness](/page/Sequential%20Compactness), after passing to a subsequence there are $u \in S^{d-1}$ and $t \in \mathbb{R}$ such that $u_n \to u$ and $t_n \to t$.[/guided]

[step:Pass the bisection inequalities to the limiting hyperplane]For $\delta>0$, define the enlarged closed positive half-space \begin{align*} C_\delta^+ := \{x \in \mathbb{R}^d : x \cdot u \geq t-\delta\} \end{align*} and the enlarged closed negative half-space \begin{align*} C_\delta^- := \{x \in \mathbb{R}^d : x \cdot u \leq t+\delta\}. \end{align*} We prove the positive inequality; the negative one is identical with the reversed sign. Fix $\delta>0$ and $\eta>0$. By tightness of $(\nu_{i,n})_{n \in \mathbb{N}}$, choose $R>0$ such that \begin{align*} \nu_{i,n}(\mathbb{R}^d \setminus B(0,R))<\eta \end{align*} for all sufficiently large $n$. Since $u_n \to u$ and $t_n \to t$, for all sufficiently large $n$ and every $x \in B(0,R)$, \begin{align*} x \cdot u_n \geq t_n \implies x \cdot u \geq t-\delta. \end{align*} Therefore \begin{align*} \{x \in \mathbb{R}^d : x \cdot u_n \geq t_n\} \subset C_\delta^+ \cup (\mathbb{R}^d \setminus B(0,R)) \end{align*} for all sufficiently large $n$. Using the bisecting property of the approximants, \begin{align*} \frac{1}{2}\nu_{i,n}(\mathbb{R}^d) \leq \nu_{i,n}(C_\delta^+) + \eta. \end{align*} Taking the upper limit along the subsequence and using weak convergence together with the [Portmanteau Theorem](/theorems/1171) for the closed set $C_\delta^+$ gives \begin{align*} \frac{1}{2}\mu_i(\mathbb{R}^d) \leq \mu_i(C_\delta^+) + \eta. \end{align*} Since $\eta>0$ was arbitrary, \begin{align*} \frac{1}{2}\mu_i(\mathbb{R}^d) \leq \mu_i(C_\delta^+). \end{align*} Letting $\delta \downarrow 0$ and using continuity from above for the finite measure $\mu_i$ yields \begin{align*} \frac{1}{2}\mu_i(\mathbb{R}^d) \leq \mu_i(\{x \in \mathbb{R}^d : x \cdot u \geq t\}). \end{align*} For the negative inequality, fix $\delta>0$ and $\eta>0$. By tightness of $(\nu_{i,n})_{n \in \mathbb{N}}$, choose $R>0$ such that \begin{align*} \nu_{i,n}(\mathbb{R}^d \setminus B(0,R))<\eta \end{align*} for all sufficiently large $n$. Since $u_n \to u$ and $t_n \to t$, for all sufficiently large $n$ and every $x \in B(0,R)$, \begin{align*} x \cdot u_n \leq t_n \implies x \cdot u \leq t+\delta. \end{align*} Therefore \begin{align*} \{x \in \mathbb{R}^d : x \cdot u_n \leq t_n\} \subset C_\delta^- \cup (\mathbb{R}^d \setminus B(0,R)) \end{align*} for all sufficiently large $n$. Using the bisecting property of the approximants, \begin{align*} \frac{1}{2}\nu_{i,n}(\mathbb{R}^d) \leq \nu_{i,n}(C_\delta^-) + \eta. \end{align*} Taking the upper limit along the subsequence and using weak convergence together with the [Portmanteau Theorem](/theorems/1171) for the closed set $C_\delta^-$ gives \begin{align*} \frac{1}{2}\mu_i(\mathbb{R}^d) \leq \mu_i(C_\delta^-) + \eta. \end{align*} Since $\eta>0$ was arbitrary, \begin{align*} \frac{1}{2}\mu_i(\mathbb{R}^d) \leq \mu_i(C_\delta^-). \end{align*} Letting $\delta \downarrow 0$ and using continuity from above for the finite measure $\mu_i$ yields \begin{align*} \frac{1}{2}\mu_i(\mathbb{R}^d) \leq \mu_i(\{x \in \mathbb{R}^d : x \cdot u \leq t\}). \end{align*} These two inequalities hold for every $i \in \{1,\dots,d\}$. Hence the affine hyperplane $\{x \in \mathbb{R}^d : x \cdot u=t\}$ bisects all the measures in the stated closed-half-space sense, completing the proof.[/step]

[guided]We now transfer the exact bisection property from the approximating measures to the limiting measure. Fix an index $i\in\{1,\dots,d\}$. For $\delta>0$, define the enlarged closed positive half-space \begin{align*} C_\delta^+ := \{x \in \mathbb{R}^d : x \cdot u \geq t-\delta\} \end{align*} and the enlarged closed negative half-space \begin{align*} C_\delta^- := \{x \in \mathbb{R}^d : x \cdot u \leq t+\delta\}. \end{align*} Why enlarge the limiting half-spaces? The sets cut out by $u_n$ and $t_n$ need not be contained in the limiting half-spaces on all of $\mathbb{R}^d$, but they are contained in small enlargements on each fixed ball once $u_n\to u$ and $t_n\to t$. Fix $\delta>0$ and $\eta>0$. By the eventual tightness argument above, applied to the weak convergence $\nu_{i,n}\rightharpoonup\mu_i$, choose $R>0$ such that \begin{align*} \nu_{i,n}(\mathbb{R}^d\setminus B(0,R))<\eta \end{align*} for all sufficiently large $n$. Since $u_n\to u$ and $t_n\to t$, for all sufficiently large $n$ and every $x\in B(0,R)$, \begin{align*} x\cdot u_n\geq t_n \implies x\cdot u\geq t-\delta. \end{align*} Therefore \begin{align*} \{x\in\mathbb{R}^d:x\cdot u_n\geq t_n\}\subset C_\delta^+\cup(\mathbb{R}^d\setminus B(0,R)) \end{align*} for all sufficiently large $n$. The approximating hyperplane bisects $\nu_{i,n}$, so \begin{align*} \frac{1}{2}\nu_{i,n}(\mathbb{R}^d)\leq\nu_{i,n}(C_\delta^+)+\eta. \end{align*} Taking the upper limit and applying the [Portmanteau Theorem](/theorems/1171) to the closed set $C_\delta^+$ gives \begin{align*} \frac{1}{2}\mu_i(\mathbb{R}^d)\leq\mu_i(C_\delta^+)+\eta. \end{align*} Since $\eta>0$ is arbitrary, \begin{align*} \frac{1}{2}\mu_i(\mathbb{R}^d)\leq\mu_i(C_\delta^+). \end{align*} Letting $\delta\downarrow0$ and using continuity from above for the finite measure $\mu_i$ yields \begin{align*} \frac{1}{2}\mu_i(\mathbb{R}^d)\leq\mu_i(\{x\in\mathbb{R}^d:x\cdot u\geq t\}). \end{align*} The negative half-space is handled by the same argument with the reversed inequality in the defining half-space. For $\delta>0$ and $\eta>0$, tightness gives $R>0$ with \begin{align*} \nu_{i,n}(\mathbb{R}^d\setminus B(0,R))<\eta \end{align*} for all sufficiently large $n$. The convergence $u_n\to u$ and $t_n\to t$ gives, for all sufficiently large $n$ and every $x\in B(0,R)$, \begin{align*} x\cdot u_n\leq t_n \implies x\cdot u\leq t+\delta. \end{align*} Hence \begin{align*} \{x\in\mathbb{R}^d:x\cdot u_n\leq t_n\}\subset C_\delta^-\cup(\mathbb{R}^d\setminus B(0,R)). \end{align*} Using the bisection property and then Portmanteau for the closed set $C_\delta^-$ gives \begin{align*} \frac{1}{2}\mu_i(\mathbb{R}^d)\leq\mu_i(C_\delta^-)+\eta. \end{align*} Letting $\eta\downarrow0$ and then $\delta\downarrow0$ gives \begin{align*} \frac{1}{2}\mu_i(\mathbb{R}^d)\leq\mu_i(\{x\in\mathbb{R}^d:x\cdot u\leq t\}). \end{align*} Thus both closed half-spaces bounded by $\{x\in\mathbb{R}^d:x\cdot u=t\}$ have at least half of the total $\mu_i$-mass. Since their union is $\mathbb{R}^d$, these inequalities are precisely the closed-half-space bisection assertion. The index $i$ was arbitrary, so the same affine hyperplane bisects all $d$ measures.[/guided]