[proofplan] We define the candidate marginal density by integrating the joint density over the $Y$-variable. Tonelli's theorem gives measurability of this candidate and identifies its total integral as the total mass of the joint density. Then, for an arbitrary Borel set $A\subset\mathbb R^m$, we compute $\mathbb P(X\in A)$ as the probability that $(X,Y)$ lies in $A\times\mathbb R^n$ and use Tonelli's theorem again to rewrite this as an integral of the candidate density over $A$. This proves that the candidate is a density of $X$, and the almost-everywhere uniqueness of Radon-Nikodym derivatives gives the final formulation for any other density $f_X$. [/proofplan]

[step:Define the marginal candidate and verify its measurability] Define the function \begin{align*} g:\mathbb R^m\to[0,\infty],\qquad g(x)=\int_{\mathbb R^n} f_{X,Y}(x,y)\,d\mathcal L^n(y). \end{align*} Since $f_{X,Y}$ is a density, it is $\mathcal B(\mathbb R^m)\otimes\mathcal B(\mathbb R^n)$-measurable and non-negative. By Tonelli's theorem for non-negative [measurable functions](/page/Measurable%20Functions) on the product [measure space](/page/Measure%20Space) $(\mathbb R^m,\mathcal B(\mathbb R^m),\mathcal L^m)\times(\mathbb R^n,\mathcal B(\mathbb R^n),\mathcal L^n)$, the section integral defining $g$ is $\mathcal B(\mathbb R^m)$-measurable. [/step]

[step:Show that the marginal candidate has total mass one] Using the standard identification $\mathcal L^{m+n}=\mathcal L^m\otimes\mathcal L^n$ on $\mathbb R^m\times\mathbb R^n$, Tonelli's theorem gives \begin{align*} \int_{\mathbb R^m} g(x)\,d\mathcal L^m(x)=\int_{\mathbb R^m}\int_{\mathbb R^n} f_{X,Y}(x,y)\,d\mathcal L^n(y)\,d\mathcal L^m(x). \end{align*} Again by Tonelli's theorem, the right-hand side equals the integral of the joint density over the product space: \begin{align*} \int_{\mathbb R^m} g(x)\,d\mathcal L^m(x)=\int_{\mathbb R^m\times\mathbb R^n} f_{X,Y}(x,y)\,d\mathcal L^{m+n}(x,y). \end{align*} Because $f_{X,Y}$ is a joint density of $(X,Y)$, this last integral is $1$. Hence \begin{align*} \int_{\mathbb R^m} g(x)\,d\mathcal L^m(x)=1. \end{align*} In particular, $g<\infty$ for $\mathcal L^m$-a.e. $x\in\mathbb R^m$, since a non-negative [measurable function](/page/Measurable%20Function) with finite integral cannot be infinite on a set of positive $\mathcal L^m$-measure. [/step]

[step:Compute probabilities of events depending only on $X$]Let $A\in\mathcal B(\mathbb R^m)$ be arbitrary. Define \begin{align*} H_A:\mathbb R^m\times\mathbb R^n\to[0,\infty),\qquad H_A(x,y)=\mathbb 1_A(x)f_{X,Y}(x,y). \end{align*} The function $H_A$ is non-negative and $\mathcal B(\mathbb R^m)\otimes\mathcal B(\mathbb R^n)$-measurable because $\mathbb 1_A:\mathbb R^m\to\{0,1\}$ is Borel-measurable and $f_{X,Y}$ is measurable. Since $(X,Y)$ has density $f_{X,Y}$, \begin{align*} \mathbb P(X\in A)=\mathbb P((X,Y)\in A\times\mathbb R^n)=\int_{A\times\mathbb R^n} f_{X,Y}(x,y)\,d\mathcal L^{m+n}(x,y). \end{align*} Using $\mathcal L^{m+n}=\mathcal L^m\otimes\mathcal L^n$ and applying Tonelli's theorem to $H_A$, we obtain \begin{align*} \int_{A\times\mathbb R^n} f_{X,Y}(x,y)\,d\mathcal L^{m+n}(x,y)=\int_{\mathbb R^m}\int_{\mathbb R^n}\mathbb 1_A(x)f_{X,Y}(x,y)\,d\mathcal L^n(y)\,d\mathcal L^m(x). \end{align*} Since $\mathbb 1_A(x)$ is independent of $y$, the inner integral is \begin{align*} \int_{\mathbb R^n}\mathbb 1_A(x)f_{X,Y}(x,y)\,d\mathcal L^n(y)=\mathbb 1_A(x)g(x). \end{align*} Therefore \begin{align*} \mathbb P(X\in A)=\int_{\mathbb R^m}\mathbb 1_A(x)g(x)\,d\mathcal L^m(x)=\int_A g(x)\,d\mathcal L^m(x). \end{align*}[/step]

[guided]We want to prove that $g$ is the density of $X$, so we must show that every Borel probability of $X$ is obtained by integrating $g$ over the corresponding Borel set. Fix an arbitrary Borel set $A\in\mathcal B(\mathbb R^m)$. The event $\{X\in A\}$ is the same as the event $\{(X,Y)\in A\times\mathbb R^n\}$, because no restriction is imposed on the $Y$-coordinate. Thus \begin{align*} \mathbb P(X\in A)=\mathbb P((X,Y)\in A\times\mathbb R^n). \end{align*} Since $(X,Y)$ has joint density $f_{X,Y}$ with respect to $\mathcal L^{m+n}$, the definition of density gives \begin{align*} \mathbb P((X,Y)\in A\times\mathbb R^n)=\int_{A\times\mathbb R^n} f_{X,Y}(x,y)\,d\mathcal L^{m+n}(x,y). \end{align*} To separate the $x$-variable from the $y$-variable, define \begin{align*} H_A:\mathbb R^m\times\mathbb R^n\to[0,\infty),\qquad H_A(x,y)=\mathbb 1_A(x)f_{X,Y}(x,y). \end{align*} This function is measurable because $\mathbb 1_A$ is Borel-measurable on $\mathbb R^m$ and $f_{X,Y}$ is product-measurable on $\mathbb R^m\times\mathbb R^n$. It is non-negative because both factors are non-negative. These are precisely the hypotheses needed to apply Tonelli's theorem for non-negative measurable functions. Using the identification $\mathcal L^{m+n}=\mathcal L^m\otimes\mathcal L^n$, Tonelli's theorem gives \begin{align*} \int_{A\times\mathbb R^n} f_{X,Y}(x,y)\,d\mathcal L^{m+n}(x,y)=\int_{\mathbb R^m}\int_{\mathbb R^n}\mathbb 1_A(x)f_{X,Y}(x,y)\,d\mathcal L^n(y)\,d\mathcal L^m(x). \end{align*} For each fixed $x\in\mathbb R^m$, the quantity $\mathbb 1_A(x)$ is constant as a function of $y$. Therefore it factors out of the inner integral: \begin{align*} \int_{\mathbb R^n}\mathbb 1_A(x)f_{X,Y}(x,y)\,d\mathcal L^n(y)=\mathbb 1_A(x)\int_{\mathbb R^n}f_{X,Y}(x,y)\,d\mathcal L^n(y). \end{align*} By the definition of $g$, this is \begin{align*} \int_{\mathbb R^n}\mathbb 1_A(x)f_{X,Y}(x,y)\,d\mathcal L^n(y)=\mathbb 1_A(x)g(x). \end{align*} Substituting this into the preceding display gives \begin{align*} \mathbb P(X\in A)=\int_{\mathbb R^m}\mathbb 1_A(x)g(x)\,d\mathcal L^m(x). \end{align*} Finally, the defining property of the indicator function converts this integral over all of $\mathbb R^m$ into an integral over $A$: \begin{align*} \mathbb P(X\in A)=\int_A g(x)\,d\mathcal L^m(x). \end{align*} This is exactly the density identity for the random vector $X$.[/guided]

[step:Conclude that $X$ is continuous and identify all versions of the density] The preceding step proves that for every $A\in\mathcal B(\mathbb R^m)$, \begin{align*} \mathbb P(X\in A)=\int_A g(x)\,d\mathcal L^m(x). \end{align*} Thus the law $\mathbb P\circ X^{-1}$ is absolutely continuous with respect to $\mathcal L^m$, and $g$ is a density of $X$. Hence $X$ is a continuous random vector. If $f_X:\mathbb R^m\to[0,\infty]$ is any other density of $X$ with respect to $\mathcal L^m$, then for every $A\in\mathcal B(\mathbb R^m)$, \begin{align*} \int_A f_X(x)\,d\mathcal L^m(x)=\mathbb P(X\in A)=\int_A g(x)\,d\mathcal L^m(x). \end{align*} By uniqueness of Radon-Nikodym derivatives up to almost-everywhere equality, $f_X=g$ $\mathcal L^m$-a.e. Therefore \begin{align*} f_X(x)=\int_{\mathbb R^n} f_{X,Y}(x,y)\,d\mathcal L^n(y) \end{align*} for $\mathcal L^m$-a.e. $x\in\mathbb R^m$. This is the marginal density formula. [/step]