The part of the FTC I am interested in says:
If $f$ is a Lebesgue-integrable function on $[a,b]$, then $F(x)=\int_a^xf(t)\,dt$ is continuous.
This is usually considered a lemma or something for the first part of the FTC, which adds the condition that $f$ is continuous and gets the extra information that $F$ is differentiable with $F'=f$. But this part of the theorem, without the continuity assumption, is giving me trouble. Combined with the fact that this is a Lebesgue integral and not a Riemann integral, it means that $f$ is not necessarily bounded, so I can't just prove that $F$ is Lipschitz-continuous as in the usual proof with Riemann-integrable functions. A simple example of a function that is not bounded but is Lebesgue-integrable on compact sets (and such that $F$ is continuous) is $f(x)=|x|^{-1/2}$.
The weakening to Lebesgue-integrable functions is my own addition, so I don't have independent confirmation that the theorem is true, but it seems clear to me that any counterexample will essentially reproduce the characteristics of the delta function, which is not a function in the usual sense. Any ideas?
Edit: It suffices to prove this for nonnegative functions $f$, because $$|F(y)-F(x)|=\left|\int_x^yf(t)\,dt\right|\le\int_x^y|f(t)|\,dt=\int_x^yg(t)\,dt=G(y)-G(x)$$ where $g(t)=|f(t)|$ and $G(x)=\int_a^xg(t)\,dt$, so if $G$ is continuous then $F$ is also continuous.
If $f$ is integrable with repect to the Lebesgue measure, then for all $\epsilon>0$, there is a $\delta>0$ such that if $mA < \delta$, then $\int_A |f| < \epsilon$.
If $f$ is bounded, this is straightforward to show. If not, let $U_M = \{ x | |f(x)| \ge M \}$ and notice that $\lim_{M \uparrow \infty} \int_{U_M} |f| = 0 $. Choose $M$ such that $\int_{U_M} |f| < { \epsilon \over 2}$,and let $\delta = {\epsilon \over 2M}$. Now suppose $m A < \delta$. Then $A = (A \cap U_M) \cup (A \setminus U_M)$. A quick calculation shows that $\int_A |f| < \epsilon$.
Now suppose $x<y$ then we have $F(y)-F(x) = \int_x^y f(x) dx$. Choose $\epsilon>0$ and let $\delta>0$ be the $\delta$ given above. Then if $|x-y| < \delta$ (meaning $m [x,y] < \delta$), we have$\int_x^y f(x) < \epsilon$. Switching the roles of $x,y$ finishes the proof.