[proofplan] The proof is an immediate unwinding of the pushforward definition of the law and the definition of absolute [continuity of measures](/theorems/1082). For a Borel null set $N$, the event $\{X\in N\}$ is precisely $X^{-1}(N)$, so its probability equals $\mu_X(N)$. Absolute continuity then forces this value to be zero. The singleton case follows by applying the first part to $N=\{a\}$, after verifying directly that $\{a\}$ is Borel and has one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) zero. [/proofplan] [step:Identify the probability of hitting a Borel set with the law of $X$] Let $N\in\mathcal B(\mathbb R)$ be such that $\mathcal L^1(N)=0$. Since $X:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is measurable, the set \begin{align*} X^{-1}(N):=\{\omega\in\Omega:X(\omega)\in N\} \end{align*} belongs to $\mathcal F$. By the definition of the pushforward measure $\mu_X=\mathbb P\circ X^{-1}$, \begin{align*} \mu_X(N)=\mathbb P(X^{-1}(N)). \end{align*} By the notation $\mathbb P(X\in N):=\mathbb P(\{\omega\in\Omega:X(\omega)\in N\})$, this gives \begin{align*} \mathbb P(X\in N)=\mu_X(N). \end{align*} [guided] Let $N\in\mathcal B(\mathbb R)$ satisfy $\mathcal L^1(N)=0$. The first point is to make the event notation explicit. Since $X:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is measurable and $N$ is Borel, the preimage \begin{align*} X^{-1}(N):=\{\omega\in\Omega:X(\omega)\in N\} \end{align*} is an element of $\mathcal F$, so its probability is defined. The law of $X$ is the pushforward measure $\mu_X=\mathbb P\circ X^{-1}$ on $(\mathbb R,\mathcal B(\mathbb R))$. Therefore, for every Borel set $B\in\mathcal B(\mathbb R)$, \begin{align*} \mu_X(B)=\mathbb P(X^{-1}(B)). \end{align*} Applying this definition to the particular Borel set $N$ gives \begin{align*} \mu_X(N)=\mathbb P(X^{-1}(N)). \end{align*} Finally, the notation $\mathbb P(X\in N)$ is shorthand for $\mathbb P(X^{-1}(N))$. Hence \begin{align*} \mathbb P(X\in N)=\mu_X(N). \end{align*} This is the bridge between the measure-theoretic hypothesis on the law and the probabilistic conclusion about the [random variable](/page/Random%20Variable). [/guided] [/step] [step:Apply absolute continuity to the Lebesgue null set] The hypothesis $\mu_X\ll\mathcal L^1$ means that for every $B\in\mathcal B(\mathbb R)$, \begin{align*} \mathcal L^1(B)=0\implies \mu_X(B)=0. \end{align*} Since $N\in\mathcal B(\mathbb R)$ and $\mathcal L^1(N)=0$, we obtain \begin{align*} \mu_X(N)=0. \end{align*} Combining this with $\mathbb P(X\in N)=\mu_X(N)$ gives \begin{align*} \mathbb P(X\in N)=0. \end{align*} [/step] [step:Deduce the singleton case from the nullity of points] Let $a\in\mathbb R$. The singleton $\{a\}$ is closed in $\mathbb R$, hence $\{a\}\in\mathcal B(\mathbb R)$. To see that it is Lebesgue null, for each $\varepsilon>0$ the interval \begin{align*} I_\varepsilon:=\left(a-\frac{\varepsilon}{2},a+\frac{\varepsilon}{2}\right) \end{align*} is an open interval containing $\{a\}$ and has Lebesgue measure \begin{align*} \mathcal L^1(I_\varepsilon)=\varepsilon. \end{align*} By monotonicity of Lebesgue measure, \begin{align*} 0\le \mathcal L^1(\{a\})\le \mathcal L^1(I_\varepsilon)=\varepsilon. \end{align*} Since this holds for every $\varepsilon>0$, we have \begin{align*} \mathcal L^1(\{a\})=0. \end{align*} Applying the first part with $N=\{a\}$ yields \begin{align*} \mathbb P(X=a)=\mathbb P(X\in\{a\})=0. \end{align*} This proves the stated particular case and completes the proof. [/step]