[proofplan] Fix a point $x\in\mathbb R$ and rewrite the empirical distribution function at $x$ as the arithmetic mean of indicator random variables. These indicators are i.i.d. Bernoulli random variables because the original observations are i.i.d., and their common expectation is exactly $F(x)$. The [strong law of large numbers](/theorems/520) then gives almost sure convergence of the empirical average to this expectation. [/proofplan] [step:Represent the empirical distribution function as an average of indicators] Fix $x\in\mathbb R$. Let $\mathcal B(\mathbb R)$ denote the Borel $\sigma$-algebra on $\mathbb R$. For each $i\in\mathbb N$, define the [random variable](/page/Random%20Variable) $Y_i:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ by \begin{align*} Y_i(\omega)=\mathbb 1_{\{X_i\le x\}}(\omega). \end{align*} The event $\{X_i\le x\}$ belongs to $\mathcal F$ because $X_i$ is measurable and $(-\infty,x]\in\mathcal B(\mathbb R)$, so each $Y_i$ is a well-defined real-valued random variable. By the definition of $F_n$, \begin{align*} F_n(x,\omega)=\frac{1}{n}\sum_{i=1}^{n}Y_i(\omega) \end{align*} for every $n\in\mathbb N$ and every $\omega\in\Omega$. [/step] [step:Verify that the indicators are i.i.d. and have mean $F(x)$] Since $X_1,X_2,\dots$ are independent and identically distributed, the random variables $Y_1,Y_2,\dots$ are also independent and identically distributed, because each $Y_i$ is obtained from $X_i$ by the same Borel map $h_x:\mathbb R\to\mathbb R$ given by \begin{align*} h_x(t)=\mathbb 1_{(-\infty,x]}(t). \end{align*} Moreover $0\le Y_i\le 1$, so $Y_i$ is integrable. Its expectation is computed by splitting over the two values of $Y_i$: \begin{align*} \mathbb E[Y_i]=1\cdot \mathbb P(Y_i=1)+0\cdot \mathbb P(Y_i=0). \end{align*} Since $Y_i=1$ exactly on the event $\{X_i\le x\}$, this becomes \begin{align*} \mathbb E[Y_i]=\mathbb P(X_i\le x)=F(x). \end{align*} [guided] The point of introducing $Y_i$ is to turn a statement about the empirical distribution function into a standard average of random variables. For the fixed number $x\in\mathbb R$, let $\mathcal B(\mathbb R)$ denote the Borel $\sigma$-algebra on $\mathbb R$, and define the map \begin{align*} Y_i:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R)). \end{align*} Its value at $\omega\in\Omega$ is \begin{align*} Y_i(\omega)=\mathbb 1_{\{X_i\le x\}}(\omega). \end{align*} This is measurable because $\{X_i\le x\}=X_i^{-1}((-\infty,x])\in\mathcal F$. Why are the $Y_i$ independent and identically distributed? Define the Borel map $h_x:\mathbb R\to\mathbb R$ by \begin{align*} h_x(t)=\mathbb 1_{(-\infty,x]}(t). \end{align*} Then $Y_i=h_x\circ X_i$ for every $i\in\mathbb N$. Applying the same measurable function to independent random variables preserves independence, and applying the same measurable function to identically distributed random variables preserves equality of distributions. Hence $Y_1,Y_2,\dots$ are i.i.d. Finally, each $Y_i$ only takes the values $0$ and $1$, so $0\le Y_i\le 1$ and $Y_i$ is integrable. Its expectation is computed from its two possible values: \begin{align*} \mathbb E[Y_i]=1\cdot \mathbb P(Y_i=1)+0\cdot \mathbb P(Y_i=0). \end{align*} The event $\{Y_i=1\}$ is the same event as $\{\mathbb 1_{\{X_i\le x\}}=1\}$, which is the same event as $\{X_i\le x\}$. Hence \begin{align*} \mathbb E[Y_i]=\mathbb P(X_i\le x)=F(x). \end{align*} Thus the empirical distribution function at $x$ is the sample mean of i.i.d. integrable random variables whose common mean is $F(x)$. [/guided] [/step] [step:Apply the strong law of large numbers] The sequence $Y_1,Y_2,\dots$ is i.i.d. and integrable, so the [strong law of large numbers](/theorems/1852) applies (citing a result not yet in the wiki: Strong Law of Large Numbers). Therefore \begin{align*} \frac{1}{n}\sum_{i=1}^{n}Y_i\xrightarrow{a.s.}\mathbb E[Y_1]. \end{align*} Using $\mathbb E[Y_1]=F(x)$ and $F_n(x,\cdot)=\frac{1}{n}\sum_{i=1}^{n}Y_i$, we obtain \begin{align*} F_n(x,\cdot)\xrightarrow{a.s.}F(x). \end{align*} Since $x\in\mathbb R$ was arbitrary, this proves the asserted pointwise almost sure convergence for every fixed $x\in\mathbb R$. [/step]