Exercise 1.6.5 (Lebesgue differentiation theorem, part I)

Question

Let $f:\mathbb{R}	o\mathbb{C}$ be an absolutely integrable function, and let $$F:\mathbb{R}	o\mathbb{C},\quad F(x) = \int_{[-\infty,x]} f \ \mathrm{d}\mu.$$ Then $F$ is continuous.

Elu Thingol · Accepted Answer

We want to demonstrate that$$\forall \epsilon>0 \quad \exists \delta > 0 \quad \forall x \in \mathbb{R}/\{x\}:\quad |x-y|\leq \delta \implies |F(x) - F(y)| \leq \epsilon.$$We analyse the last term. We have, by additivity of the integrals, 
ewline   \begin{align*}      \left|F(y) - F(x) ight| &=\left|\int_{[-\infty,y]}f\,\mathrm{d}\mu - \int_{[-\infty,x]}f\,\mathrm{d}\mu ight| 
ewline= \left| \int_{[y,x]\cup [x,y]}f\,\mathrm{d}\mu ight| 
ewline\end{align*}By Theorem 1.3.20 (Approximation of $L^1$ functions) and by triangle inequality (Lemma 1.3.19) we can find a compactly supported continuous function $g$ such that $|\int f| - |\int g| \leq \int |f-g| \leq 1/2\epsilon$. We thus continue our inequality chain\begin{align*}= \left| \int_{[y,x]\cup [x,y]}f\,\mathrm{d}\mu ight|
ewline&\leq \frac{1}{2}\epsilon + \left| \int_{[y,x]\cup [x,y]}g\,\mathrm{d}\mu ight|
ewline\end{align*}Since a continuous function on a compact interval is bounded, $g$ must have a bound on $\forall z\in [x,y]\cup [y,x]: |f(x)|\leq M$. Thus,\begin{align*}= \frac{1}{2}\epsilon + \left| \int_{[y,x]\cup [x,y]}g\,\mathrm{d}\mu ight|
ewline&\leq \frac{1}{2}\epsilon + \left|Might| \cdot \left|x-yight|
ewline\end{align*}Setting $\delta_\epsilon := \dfrac{\epsilon}{2M}$, we obtain the desired result.

Exercise 1.6.5 (Lebesgue differentiation theorem, part I)

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Exercise 1.6.5 (Lebesgue differentiation theorem, part I)

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