[proofplan] The interval identity follows by writing the event $\{a<X\le b\}$ as a set difference between two sublevel events of $X$. The density formula is then the defining property of a Lebesgue density applied to the Borel set $(a,b]$. Continuity of $F_X$ follows from the absolute continuity of Lebesgue integration for the integrable density $f_X$. For differentiability, replace $f_X$ by the a.e.-equal representative $\widetilde f_X$ on small intervals around $x$ and show that the relevant difference quotients are averages of a [continuous function](/page/Continuous%20Function) over shrinking intervals. [/proofplan]

[step:Identify the interval event as a difference of sublevel events]Fix $a,b\in\mathbb R$ with $a\le b$. Define the events \begin{align*} A_t:=\{\omega\in\Omega:X(\omega)\le t\}\in\mathcal F \end{align*} for $t\in\mathbb R$. Since $a\le b$, we have $A_a\subset A_b$, and \begin{align*} A_b\setminus A_a=\{\omega\in\Omega:a<X(\omega)\le b\}. \end{align*} By finite additivity of $\mathbb P$ on the disjoint union $A_b=A_a\cup(A_b\setminus A_a)$, \begin{align*} \mathbb P(a<X\le b)=\mathbb P(A_b)-\mathbb P(A_a)=F_X(b)-F_X(a). \end{align*}[/step]

[guided]Fix $a,b\in\mathbb R$ with $a\le b$. The distribution function is defined in terms of events of the form $\{X\le t\}$, so we compare the two sublevel events at $a$ and $b$. For each $t\in\mathbb R$, define \begin{align*} A_t:=\{\omega\in\Omega:X(\omega)\le t\}. \end{align*} Because $X:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is measurable and $(-\infty,t]\in\mathcal B(\mathbb R)$, the set $A_t=X^{-1}((-\infty,t])$ belongs to $\mathcal F$. Since $a\le b$, every outcome satisfying $X(\omega)\le a$ also satisfies $X(\omega)\le b$, so $A_a\subset A_b$. The part of $A_b$ not already contained in $A_a$ is exactly the event where $X$ is at most $b$ but not at most $a$: \begin{align*} A_b\setminus A_a=\{\omega\in\Omega:a<X(\omega)\le b\}. \end{align*} The union $A_b=A_a\cup(A_b\setminus A_a)$ is disjoint. Therefore finite additivity of the [probability measure](/page/Probability%20Measure) gives \begin{align*} \mathbb P(A_b)=\mathbb P(A_a)+\mathbb P(A_b\setminus A_a). \end{align*} Rearranging and using $F_X(t)=\mathbb P(A_t)$ yields \begin{align*} \mathbb P(a<X\le b)=F_X(b)-F_X(a). \end{align*}[/guided]

[step:Apply the density property to the Borel interval $(a,b]$] The set $(a,b]$ belongs to $\mathcal B(\mathbb R)$. Since $f_X$ is a Lebesgue density for $X$, the defining density identity gives \begin{align*} \mathbb P(a<X\le b)=\mathbb P(X\in(a,b])=\int_{(a,b]} f_X(u)\,d\mathcal L^1(u). \end{align*} [/step]

[step:Use absolute continuity of Lebesgue integration to prove continuity of $F_X$] For each $t\in\mathbb R$, the density property applied to $(-\infty,t]\in\mathcal B(\mathbb R)$ gives \begin{align*} F_X(t)=\int_{(-\infty,t]} f_X(u)\,d\mathcal L^1(u). \end{align*} [claim:Integrals of $L^1$ functions are small on sets of small measure] Let $g:\mathbb R\to[0,\infty)$ be $\mathcal B(\mathbb R)$-measurable and satisfy \begin{align*} \int_{\mathbb R} g(u)\,d\mathcal L^1(u)<\infty. \end{align*} For every $\varepsilon>0$ there exists $\delta>0$ such that, whenever $E\in\mathcal B(\mathbb R)$ and $\mathcal L^1(E)<\delta$, \begin{align*} \int_E g(u)\,d\mathcal L^1(u)<\varepsilon. \end{align*} [/claim] [proof] Fix $\varepsilon>0$. For each $c>0$, define the Borel set \begin{align*} E_c:=\{u\in\mathbb R:g(u)>c\}. \end{align*} Let $\mathbb{1}_{E_c}:\mathbb R\to\{0,1\}$ denote the indicator function of $E_c$, defined by $\mathbb{1}_{E_c}(u)=1$ for $u\in E_c$ and $\mathbb{1}_{E_c}(u)=0$ for $u\notin E_c$. Define the [measurable function](/page/Measurable%20Function) $h_c:\mathbb R\to[0,\infty)$ by $h_c(u)=g(u)\mathbb{1}_{E_c}(u)$. If $(c_n)_{n=1}^{\infty}$ is any sequence in $(0,\infty)$ with $c_n\to\infty$, then $h_{c_n}(u)\to0$ for every $u\in\mathbb R$, and $0\le h_{c_n}\le g$. Since \begin{align*} \int_{\mathbb R} g(u)\,d\mathcal L^1(u)<\infty, \end{align*} the [dominated convergence theorem](/theorems/4), applied to the sequence $(h_{c_n})_{n=1}^{\infty}$ dominated by $g$, gives \begin{align*} \lim_{n\to\infty}\int_{E_{c_n}} g(u)\,d\mathcal L^1(u)=0. \end{align*} Because this holds for every sequence $c_n\to\infty$, we have \begin{align*} \lim_{c\to\infty}\int_{E_c} g(u)\,d\mathcal L^1(u)=0. \end{align*} Choose $c>0$ such that \begin{align*} \int_{\{u\in\mathbb R:g(u)>c\}} g(u)\,d\mathcal L^1(u)<\frac{\varepsilon}{2}. \end{align*} Define \begin{align*} \delta:=\frac{\varepsilon}{2c}. \end{align*} If $E\in\mathcal B(\mathbb R)$ and $\mathcal L^1(E)<\delta$, then split $E$ into $E\cap\{g\le c\}$ and $E\cap\{g>c\}$. By monotonicity of the [Lebesgue integral](/page/Lebesgue%20Integral), \begin{align*} \int_E g(u)\,d\mathcal L^1(u)\le c\,\mathcal L^1(E)+\int_{\{u\in\mathbb R:g(u)>c\}} g(u)\,d\mathcal L^1(u). \end{align*} Using the choices of $\delta$ and $c$ gives \begin{align*} \int_E g(u)\,d\mathcal L^1(u)<c\delta+\frac{\varepsilon}{2}=\varepsilon. \end{align*} [/proof] Fix $t_0\in\mathbb R$ and $\varepsilon>0$. Taking $E=\mathbb R$ in the density identity gives \begin{align*} \int_{\mathbb R} f_X(u)\,d\mathcal L^1(u)=\mathbb P(X\in\mathbb R)=1, \end{align*} so $f_X$ satisfies the hypothesis of the claim. Apply the claim with $g=f_X$ to obtain $\delta>0$ such that every Borel set $E\subset\mathbb R$ with $\mathcal L^1(E)<\delta$ satisfies \begin{align*} \int_E f_X(u)\,d\mathcal L^1(u)<\varepsilon. \end{align*} If $t\in\mathbb R$ and $|t-t_0|<\delta$, then the symmetric difference of $(-\infty,t]$ and $(-\infty,t_0]$ is an interval of [Lebesgue measure](/page/Lebesgue%20Measure) $|t-t_0|$. Hence \begin{align*} |F_X(t)-F_X(t_0)|\le \int_{(\min\{t,t_0\},\max\{t,t_0\}]} f_X(u)\,d\mathcal L^1(u)<\varepsilon. \end{align*} Thus $F_X$ is continuous at $t_0$. Since $t_0$ was arbitrary, $F_X$ is continuous on $\mathbb R$. [/step]

[step:Compute the derivative from the continuous density representative]Fix $x\in\mathbb R$, and assume $\widetilde f_X:\mathbb R\to[0,\infty)$ is $\mathcal B(\mathbb R)$-measurable, satisfies $\widetilde f_X=f_X$ $\mathcal L^1$-a.e., and is continuous on an open neighbourhood of $x$. Choose $r>0$ such that $\widetilde f_X$ is continuous on $(x-r,x+r)$. Let $h\in\mathbb R$ satisfy $0<|h|<r$. Since $f_X=\widetilde f_X$ $\mathcal L^1$-a.e., their integrals over every Borel subset of $(x-r,x+r)$ agree. If $h>0$, then \begin{align*} \frac{F_X(x+h)-F_X(x)}{h}=\frac{1}{h}\int_{(x,x+h]} \widetilde f_X(u)\,d\mathcal L^1(u). \end{align*} Therefore \begin{align*} \left|\frac{F_X(x+h)-F_X(x)}{h}-\widetilde f_X(x)\right|\le \sup_{u\in(x,x+h]}|\widetilde f_X(u)-\widetilde f_X(x)|. \end{align*} If $h<0$, then \begin{align*} \frac{F_X(x+h)-F_X(x)}{h}=\frac{1}{|h|}\int_{(x+h,x]} \widetilde f_X(u)\,d\mathcal L^1(u), \end{align*} and hence \begin{align*} \left|\frac{F_X(x+h)-F_X(x)}{h}-\widetilde f_X(x)\right|\le \sup_{u\in(x+h,x]}|\widetilde f_X(u)-\widetilde f_X(x)|. \end{align*} By continuity of $\widetilde f_X$ at $x$, both suprema tend to $0$ as $h\to0$. Hence the two-sided derivative exists and \begin{align*} F_X'(x)=\widetilde f_X(x). \end{align*}[/step]

[guided]Fix $x\in\mathbb R$. We are given a $\mathcal B(\mathbb R)$-measurable representative $\widetilde f_X:\mathbb R\to[0,\infty)$ such that $\widetilde f_X=f_X$ $\mathcal L^1$-a.e. and such that $\widetilde f_X$ is continuous on an open neighbourhood of $x$. Choose $r>0$ with the property that $\widetilde f_X$ is continuous on the interval $(x-r,x+r)$. The point of using $\widetilde f_X$ is that the derivative of $F_X$ is a pointwise statement, while a density is only determined up to $\mathcal L^1$-a.e. equality. Since $f_X=\widetilde f_X$ $\mathcal L^1$-a.e., the two functions have the same Lebesgue integral over every Borel set. Thus for every Borel set $E\subset(x-r,x+r)$, \begin{align*} \int_E f_X(u)\,d\mathcal L^1(u)=\int_E \widetilde f_X(u)\,d\mathcal L^1(u). \end{align*} Let $h\in\mathbb R$ satisfy $0<|h|<r$. First suppose $h>0$. The increment of the distribution function over $[x,x+h]$ is the probability of the interval event $(x,x+h]$, so the density formula gives \begin{align*} F_X(x+h)-F_X(x)=\int_{(x,x+h]} \widetilde f_X(u)\,d\mathcal L^1(u). \end{align*} Dividing by $h$ gives \begin{align*} \frac{F_X(x+h)-F_X(x)}{h}=\frac{1}{h}\int_{(x,x+h]} \widetilde f_X(u)\,d\mathcal L^1(u). \end{align*} Subtract $\widetilde f_X(x)$, written as the average of the constant function $\widetilde f_X(x)$ over the interval $(x,x+h]$: \begin{align*} \widetilde f_X(x)=\frac{1}{h}\int_{(x,x+h]} \widetilde f_X(x)\,d\mathcal L^1(u). \end{align*} The triangle inequality and monotonicity of the Lebesgue integral yield \begin{align*} \left|\frac{F_X(x+h)-F_X(x)}{h}-\widetilde f_X(x)\right|\le \frac{1}{h}\int_{(x,x+h]}|\widetilde f_X(u)-\widetilde f_X(x)|\,d\mathcal L^1(u). \end{align*} The average of a nonnegative function is bounded above by its supremum on the interval, so \begin{align*} \left|\frac{F_X(x+h)-F_X(x)}{h}-\widetilde f_X(x)\right|\le \sup_{u\in(x,x+h]}|\widetilde f_X(u)-\widetilde f_X(x)|. \end{align*} Now suppose $h<0$. Then $x+h<x$, and the same density formula gives \begin{align*} F_X(x)-F_X(x+h)=\int_{(x+h,x]} \widetilde f_X(u)\,d\mathcal L^1(u). \end{align*} Since $h<0$, dividing $F_X(x+h)-F_X(x)$ by $h$ changes the sign in both numerator and denominator: \begin{align*} \frac{F_X(x+h)-F_X(x)}{h}=\frac{1}{|h|}\int_{(x+h,x]} \widetilde f_X(u)\,d\mathcal L^1(u). \end{align*} Repeating the same average argument on $(x+h,x]$ gives \begin{align*} \left|\frac{F_X(x+h)-F_X(x)}{h}-\widetilde f_X(x)\right|\le \sup_{u\in(x+h,x]}|\widetilde f_X(u)-\widetilde f_X(x)|. \end{align*} Because $\widetilde f_X$ is continuous at $x$, for every $\varepsilon>0$ there exists $\eta\in(0,r)$ such that $|u-x|<\eta$ implies \begin{align*} |\widetilde f_X(u)-\widetilde f_X(x)|<\varepsilon. \end{align*} If $0<|h|<\eta$, then the relevant interval, either $(x,x+h]$ or $(x+h,x]$, lies inside $(x-\eta,x+\eta)$. Hence the corresponding supremum is at most $\varepsilon$. Therefore \begin{align*} \lim_{h\to0}\frac{F_X(x+h)-F_X(x)}{h}=\widetilde f_X(x), \end{align*} which proves that $F_X$ is differentiable at $x$ and that $F_X'(x)=\widetilde f_X(x)$.[/guided]