$F(x)=\int^{x}_{a} f(y) dy$ continuous (Lebesgue Integral)

Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be Lebesgue integrable and let $a \in \mathbb{R}$. We wish to show that $F(x)=\int^{x}_{a} f(y) dy$ continuous.

I know, of course, what we need. We need to choose $x,y$ (without loss of generality, assume $x<y$) such that $$|\int^{y}_{a}{f} - \int^{x}_{a}{f}|< \epsilon.$$ Because $|\int^{y}_{a}{f} - \int^{x}_{a}{f}|=|\int^{y}_{x}{f}| \leq \int_{[x,y]}|f|$, our problem requires us to make $\int_{[x,y]}|f| < \epsilon$.

This is where I'm stuck. In fact, there are several problem similar to this on which I'm stuck. Clearly, there is something I'm missing or don't understand. (I have been studying for Qualifying Exams and I'm literally getting tripped up on every question.) Advice and help would be great. Perhaps, something expository concerning making integrals like the one above as small as we like would be remedy some of the other issue I'm having as well. Thank you!

*I went back and proved the analogous problem for the Riemann Integral. It was quite simple because if $f$ is Riemann integrable, then $f$ is bounded. I thought perhaps I could generalize from this, but I am not making the connection.

Solution Attempt: We make the following observations. Because $f$ is integrable, $\int {|f|} < \infty$ (and so is $\int_{[x,y]}{|f|}$ by monotonicity) . By definition, for every $\epsilon > 0$ we can find a simple function $\psi$ such that $\int_{[x,y]}{|f|} - \frac{\epsilon}{2} < \int{\psi} \leq \int_{[x,y]}{|f|}$. Let $\psi = \sum^{n}_{i=1} a_{i} \chi_{E_{i}}$. Notice that $\psi \leq f\chi_{[x,y]}$, $\psi = \psi \chi_{[x,y]}$ and $$\int{\psi} = \int_{[x,y]}{\psi}= \sum^{n}_{i=1} a_{i} m(E_{i}) \leq \sum^{n}_{i=1} a_{i} m([x,y]) = m([x,y]) \sum^{n}_{i=1} a_{i}. $$

The preceding paragraph suggests we take $$\delta = \frac{\epsilon}{2 \sum^{n}_{i=1} a_{i}}.$$ Then, if $m([x,y]) \leq \delta$, $$\int_{[x,y]} {|f|} = \int_{[x,y]} {|f|} - \int{\psi} + \int{\psi} < \frac{\epsilon}{2} + m([x,y])\sum^{n}_{i=1} a_{i} .$$ Then, the result follows from out choice of $\delta$.

Answer 1

Here is a slightly more general approach for comparison.

If $f$ is integrable on [a,b], then for any $\epsilon > 0$ there exists $\delta > 0$ such that for any measurable set $D \subset [a,b]$ with measure $\mu(D) < \delta$, then

$$\left|\int_{D}f(x)\,dx\right|< \epsilon.$$

Since $f$ is measurable , then for any $\epsilon>0$ it can be approximated by a simple function $\psi$ such that

$$\int_{a}^{b}|f(x)-\psi(x)|\,dx < \epsilon /2.$$

(To show this apply to $f^+$ and $f^-$ the theorem that any non-negative measurable function is the limit of a monotonically increasing sequence of simple functions.)

A simple function is bounded so let $M$ be an upper bound for $|\psi (x)|$ on $D$. Choose $\delta = \epsilon/2M$.

Then

$$\left|\int_{D}f(x)\,dx\right| \leq \int_{D}|\psi(x)|\,dx+\int_{D}|f(x)-\psi(x)|\,dx \leq \\\int_{D}|\psi(x)|\,dx+\int_{a}^{b}|f(x)-\psi(x)|\,dx< M \mu(D)+\epsilon/2<\epsilon.$$

In your case use $D = [x,y]$.