The following comes from Bass' book on Real Analysis: (Here $dy$ is Lebesgue measure)
Exercise 7.6
Suppose $f:\mathbb{R}\to\mathbb{R}$ is integrable, $a\in \mathbb{R}$, and we define $F(x):=\int_a^x f(y)\, dy$. Show that $F$ is a continuous function.
Another answer on stack exchange covers the case where $f$ is assumed to be Riemann integrable, with $f$ and $F$ being functions from some $[a,b]$. The proof there uses the fact that $f$ is Riemann integrable if and only if it is bounded and continuous almost everywhere.
Folland's book on analysis appears to define integration for functions $f:X\to[-\infty,\infty]$ on the extended real numbers. ("The integral defined in the previous section [for functions $f:\mathbb{R}\to[0,\infty]$] can be extended to real-valued measurable functions $f$…" Except, real-valued means $\mathbb{R}$ but extending the old definition forces the inclusion of infinities. However, as far as I can tell the actual definition as integrating the positive and negative parts agrees with taking $[-\infty,\infty]$-valued functions.
I only bring this up because it seems exercise 7.6 should hold for: $f:\mathbb{R}\to[-\infty,\infty]$. $$f(x):=\begin{cases} x^{-1/2}& x>0\\ (-x)^{-1/2} &x<0 \\ \infty & x=0\end{cases}$$
In trying a few things I also wondered if the $F$ from 7.6 is always measurable. (this should be an easy yes?)
It's worth noting that chapter 7 of Bass covers the usual monotone, dominated, etc. theorems.
The question then is if the modified exercise is correct, how different are the lebesgue measure -vs- riemann measure proofs, how to go about either version (Lebesgue preferred for the modified exercise)
Best Answer
I believe the proof is somewhat different than the case of Riemann integration. Also, as a comment noted, allowing functions to take values $\pm \infty$ is harmless (an integrable function can only take these values on a measure zero set). All integrable functions are measurable; this may be easy or may take some work, depending on exact definitions.
One proof of the main exercise uses the dominated convergence theorem. Fix $x_0$. For convenience, let $\chi_{(x_0,x)}$ denote the characteristic function of the interval when $x_0 < x$, and denote $-\chi_{(x,x_0)}$ otherwise. Then $$\lim_{x \to x_0} F(x) - F(x_0) = \lim_{x \to x_0} \int_{x_0}^x f(y)dy = \lim_{x\to x_0} \int_{-\infty}^{\infty} \chi_{(x_0,x)}f(y)dy = \int_{-\infty}^{\infty}\lim_{x\to x_0} \chi_{(x_0,x)}f(y)dy = \int_{-\infty}^{\infty} 0 dy = 0\text{.}$$