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Let $\lambda$ be the Lebesgue measure on $[a, b]$ and $f:[a, b] \to \mathbb{R}$ $\lambda$-integrable. We define $F:[a, b] \to \mathbb R$ by $F(x) := \int_a^x f \mathrm d \lambda$ for all $x \in [a, b]$. I'm reading a proof of Lebesgue differentiation theorem from this note.

Theorem $F$ is differentiable $\lambda$-a.e. and $F' = f$ $\lambda$-a.e.

The proof starts by considering a bounded non-negative $f$. Then the proof extends to unbounded non-negative $f$. A crucial step in the bootstrap is the identity $$ \int_a^x F' \mathrm d \lambda = F(x)-F(a). $$

This identity is obtained in the first case because $f$ is bounded by some constant $M>0$. Could you explain how it is obtained in the case $f$ is unbounded?

Proof Let's prove that $F$ is of bounded variation. Let $a =x_1 \le x_2 \le \cdots \le x_n = b$. Then $$ \sum_{i=1}^{n-1} \left| F (x_{i+1}) - F(x_{i}) \right| = \sum_{i=1}^{n-1}\left|\int_{x_{i}}^{x_{i+1}} f \mathrm d \lambda \right| \le \sum_{i=1}^{n-1} \int_{x_{i}}^{x_{i+1}} |f| \mathrm d \lambda = \|f\|_{L_1(\lambda)}. $$

Hence $\operatorname{Var}_{[a, b]} F \leq \|f\|_{L_1(\lambda)} <\infty$. It follows that $F$ is differentiable $\lambda$-a.e. and $$ \int_a^x |F'| \mathrm d \lambda \leq \operatorname{Var}_{[a, x]} F \quad \forall x \in [a, b]. $$

  1. We consider the case there is $M>0$ such that $0 \le f \le M$. Then $F$ is increasing. Hence $F' \ge 0$ and $\operatorname{Var}_{[a, x]} F = F(x)-F(a)$ for all $x \in [a, b]$. It follows that $$ \int_a^x F' \mathrm d \lambda \leq F(x)-F(a) \quad \forall x \in [a, b]. $$

Let $g:= M-f$. Then $0 \le g \le M$. We define $G:[a, b] \to \mathbb R$ by $G(x) := \int_a^x g \mathrm d \lambda$ for $x \in [a, b]$. Then $G (x)= M(x-a)- F(x)$ for $x \in [a, b]$. As for $F$, we have $G$ is differentiable $\lambda$-a.e. and $$ \int_a^x G' \mathrm d \lambda \le G(x)-G(a) \quad \forall x \in [a, b]. $$

First, it follows from $G' = M-F'$ $\lambda$-a.e that $\int_a^x G' \mathrm d \lambda = M(x-a) - \int_a^x F' \mathrm d \lambda$. Second, $G(x)-G(a) = M(x-a)- (F(x) - F(a))$. Then for $x \in [a, b]$, $$ \begin{align} M(x-a) - \int_a^x F' \mathrm d \lambda &\le M(x-a)- (F(x) - F(a)) \\ &\le M(x-a)- \int_a^x F' \mathrm d \lambda. \end{align} $$

It follows that $$ \int_a^x F' \mathrm d \lambda = F(x)-F(a) = \int_a^x f \mathrm d \lambda\quad \forall x \in [a, b]. $$

By this lemma, $F' = f$ $\lambda$-a.e.

  1. We consider the case $f \ge 0$. Let $f_n := f \wedge n$. Then $f_n \uparrow f$ everywhere. We define $$ F_n :[a, b] \to \mathbb R, x \mapsto \int_a^x f_n \mathrm d \lambda. $$

By monotone convergence theorem, $F_n \uparrow F$ everywhere. Then $$ \begin{align} \int_a^x F' \mathrm d \lambda &\overset{\color{red}{???}}{=} F(x)-F(a) \\ &= \lim_n (F_n(x)-F_n(a)) \\ &= \lim_n \int_a^x f_n \mathrm d \lambda \quad \text{by 1.}\\ &= \int_a^x f \mathrm d \lambda \quad \text{by monotone convergence theorem}. \end{align} $$

It follows that $F' = f$ $\lambda$-a.e.

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1 Answers1

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I have found a fix from this thread.

  • Second, we assume $f \ge 0$. We define $$ f_n(x) := \begin{cases} f(x) & \text{if } f(x) \le n, \\ 0 & \text {otherwise}. \end{cases} $$

Let $g_n := f - f_n$. Then $f = f_n + g_n$ and thus $F = F_n + G_n$ with $$ F_n (x):=\int_a^x f_n \mathrm d \lambda \quad \text{and} \quad G_n (x):=\int_a^x g_n \mathrm d \lambda \quad \forall x \in [a, b]. $$

Because $|f_n| \le n$, we get $F'_n = f_n$ $\lambda$-a.e. It follows from $g_n \ge 0$ that $G_n$ is increasing and thus $G'_n \ge 0$ $\lambda$-a.e. So $$ F'=F_n'+G'_n \ge F'_n = f_n \quad \lambda \text{-a.e}. $$

Taking the limit, we get $F' \ge f$ $\lambda$-a.e. and thus $$ \int_a^x F' \mathrm d \lambda \ge \int_a^x f \mathrm d \lambda = F(x)-F(a) \quad \forall x \in [a, b]. $$

On the other hand, we have $$ \int_a^x F' \mathrm d \lambda \le \operatorname{Var}_{[a, x]} F = F(x)- F(a) \quad \forall x \in [a, b]. $$

It follows that $$ \int_a^x F' \mathrm d \lambda = F(x)-F(a) \quad \forall x \in [a, b]. $$

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