The Derivative of Functions of Lebesgue Integrals

Recall from the Integral Criteria for Functions to be Zero Almost Everywhere page the following results:

• 1. If $f$ is a nonnegative Lebesgue integrable function defined on a Lebesgue measurable set $E$ and $\displaystyle{\int_E f = 0}$ then $f(x) = 0$ a.e. on $E$.
• 2. If $f$ is a Lebesgue integrable function on $[a, b]$ and $\displaystyle{F(x) = \int_a^x f(t) \: dt}$ then if $F(x) = 0$ for all $x \in [a, b]$ then $f(x) = 0$ a.e. on $E$.

We now prove a very important result

 Theorem 1: Let $f$ be a bounded Lebesgue integrable function on $[a, b]$ and let $\displaystyle{F(x) = \int_a^x f(t) \: dt}$. Then $F'(x) = f(x)$ almost everywhere on $[a, b]$.
• Proof: From the theorem stated above, since $f$ is Lebesgue integrable on $[a, b]$ we have that $F$ is of bounded variation on $[a, b]$. Recall from the Functions of Bounded Variation on Closed Intervals page that then this means that $F$ can be expressed as the difference of two increasing functions on $[a, b]$, say $F = F_1 - F_2$ where $F_1$ and $F_2$ are increasing on $[a, b]$. From Lebesgue's Theorem for the Differentiability of Monotone Functions, since $F_1$ and $F_2$ are both monotone functions, $F_1$ and $F_2$ are differentiable almost everywhere on $(a, b)$ and so $F = F_1 - F_2$ is differentiable almost everywhere on $[a, b]$.
• Now since $f$ is bounded on $[a, b]$ there exists an $M \in \mathbb{R}$, $M > 0$ such that for all $x \in [a, b]$ we have that:
(1)
• For each $n \in \mathbb{N}$, define a sequence of functions on $[a, b]$ by:
(2)
\begin{align} \quad F_n(x) = \frac{F \left ( x + \frac{1}{n} \right ) - F(x)}{\frac{1}{n}} \end{align}
• By convention, we will let $F(x) = F(b)$ for all $x \geq b$. Since we have already established that $F$ is continuous on $[a, b]$ we have that each $F_n$ is a Lebesgue measurable function on $[a, b]$. So $(F_n(x))_{n=1}^{\infty}$ is a sequence of Lebesgue measurable functions. Furthermore, for each $n \in \mathbb{N}$ we have that:
(3)
\begin{align} \quad | F_n(x) | &= \biggr \lvert \frac{F \left ( x + \frac{1}{n} \right ) - F(x)}{\frac{1}{n}} \biggr \rvert \\ &= n \biggr \lvert \int_a^{x + \frac{1}{n}} f(t) \: dt - \int_a^{x} f(t) \: dt \biggr \rvert \\ &= n \biggr \lvert \int_{x}^{x + \frac{1}{n}} f(t) \: dt \biggr \rvert \\ & \leq n \biggr \lvert \int_{\left [ x, x + \frac{1}{n} \right ]} f(t) \: dt \biggr \rvert \\ & \leq n \int_{\left [ x, x + \frac{1}{n} \right ]} |f(t)| \: dt \\ & \leq n l\left ( \left [x, x + \frac{1}{n} \right ] \right ) M \\ & \leq n \cdot \frac{1}{n} M \\ & \leq M \end{align}
(4)
\begin{align} \quad \int_a^b F'(x) \: dx = \lim_{n \to \infty} \int_a^b F_n(x) \end{align}
• And similarly, for every $c \in [a, b]$ we have that:
(5)
\begin{align} \quad \int_a^c F'(x) \: dx & = \lim_{n \to \infty} \int_a^c F_n(x) \: dx \\ &= \lim_{n \to \infty} \int_a^c \frac{F \left ( x + \frac{1}{n} \right ) - F(x)}{\frac{1}{n}} \: dx \\ &= \lim_{n \to \infty} n \left [ \int_a^c F \left ( x + \frac{1}{n} \right ) \: dx - \int_a^c F(x) \: dx \right ] \\ &= \lim_{n \to \infty} n \left [ \int_{a + \frac{1}{n}}^{c + \frac{1}{n}} F(x) \: dx - \int_a^c F(x) \: dx \right ] \\ &= \lim_{n \to \infty} n \left [ \int_c^{c + \frac{1}{n}} F(x) \: dx - \int_a^{a+\frac{1}{n}} F(x) \: dx \right ] \\ &= F(c) - F(a) \\ &= F(c) \\ &= \int_a^c f(x) \: dx \end{align}
• So for all $c \in [a, b]$ we have that:
(6)
\begin{align} \quad \int_a^c F'(x) \: dx = \int_a^c f(x) \: dx \end{align}
• Therefore, for all $c \in [a, b]$ we have that:
(7)
\begin{align} \quad \int_a^c [F'(x) - f(x)] \: dx = 0 \end{align}
• By the second result (2) mentioned at the top of the page we have that $F'(x) - f(x) = 0$ almost everywhere on $[a, b]$, so $F'(x) = f(x)$ almost everywhere on $[a, b]$. $\blacksquare$
 Theorem 2: Let $f$ be Lebesgue integrable on $[a, b]$ and let $\displaystyle{F(x) = \int_a^x f(t) \: dt}$. Then $F'(x) = f(x)$ almost everywhere on $[a, b]$.
• Proof: Assume that $f$ is a nonnegative function on $[a, b]$. Then $F$ is an increasing function on $[a, b]$. Furthermore, since $f$ is a Lebesgue integrable function we have that $F$ is of bounded variation on $[a, b]$ and so $F$ is differentiable almost everywhere on $[a, b]$ and by Lebesgue's Theorem for the Differentiability of Monotone Functions we have that $F'(x) \geq 0$ wherever $F'$ exists. For each $n \in \mathbb{N}$ let:
(8)
\begin{align} \quad f_n(x) = \min \{ f(x), n \} \end{align}
• Then each $f_n$ is a bounded Lebesgue integrable function on $[a, b]$ and furthermore, $f(x) - f_n(x) \geq 0$ on $[a, b]$. So by the previous theorem we have that for each $n \in \mathbb{N}$ and for almost every $x \in [a, b]$:
(9)
\begin{align} \quad \frac{d}{dx} \int_a^x f_n(x) \: dx = f_n(x) \quad (*) \end{align}
• For each $n \in \mathbb{N}$ let:
(10)
\begin{align} \quad G_n(x) = \int_a^x [f(t) - f_n(t)] \: dt \end{align}
• Then each $G_n$ is an increasing function. Furthermore, $G_n'$ exists almost everywhere on $[a, b]$ (since $f'$ and $f_n'$ exists almost everywhere on $[a, b]$) and $G_n'(x) \geq 0$ almost everywhere on $[a, b]$. Now:
(11)
\begin{align} \quad F(x) &= \int_a^x f(t) \: dt \\ &= \int_a^x [f(t) - f_n(t)] \: dt + \int_a^x f_n(t) \: dt \\ &= G_n(x) + \int_a^x f_n(t) \: dt \end{align}
• We differentiable both sides of the equation above with respect to $x$ to get:
(12)
\begin{align} \quad F'(x) &= G_n'(x) + \frac{d}{dx} \int_a^x f_n(t) \: dt \\ & \overset{(*)} = G_n'(x) + f_n(x) \end{align}
• Since $G_n'(x) \geq 0$ almost everywhere on $[a, b]$ we have that $F'(x) \geq f_n(x)$ almost everywhere on $[a, b]$ and so for all $n \in \mathbb{N}$ we have that $F'(x) \geq f(x)$ almost everywhere on $[a, b]$, so:
(13)
\begin{align} \quad \int_a^b F'(x) \: dx \geq \int_a^b f(x) \: dx = \int_a^b f(t) \: dt - \int_a^a f(t) \: dt = F(b) - F(a) \\ \end{align}
(14)
\begin{align} \quad \int_a^b F'(x) \: dx \leq F(b) - F(a) \end{align}
• Therefore:
(15)
\begin{align} \quad \int_a^b F'(x) \: dx &= F(b) - F(a) \\ &= \int_a^b f(x) \: dx \end{align}
• So $\displaystyle{\int_a^b [F'(x) - f(x)] \: dx = 0}$ and $F'(x) - f(x) \geq 0$, so by the result (1) at the top of this page we have that $F'(x) = f(x)$ almost everywhere on $[a, b]$.
• Now in general, if $f$ is any Lebesgue integrable function on $[a, b]$ then we've already proven that $F$ is of bounded variation on $[a, b]$. So $F$ is the difference of two increasing functions on $[a, b]$. So $F'(x) = f(x)$ almost everywhere on $[a, b]$. $\blacksquare$