Function with compact support has uniformly continuous integral over $(-\infty,x]$? If $f$ is a bounded Lebesgue measurable function that has compact support, then is $\int_{-\infty}^x f(t)dt$ uniformly continuous?
I have just learned of compact support and do not see how compact support can lead to uniform continuity since the integral here is not over a compact set? I believe this is a condition for a distribution function in probability and would like to understand why. Also, what if the function $f$ was just integrable, then would this integral still be uniformly continuous?
 A: Hint:
First, show that the function is continuous. Then, prove the following.

Let $g:\mathbb R\to\mathbb R$ be continuous. Suppose $\displaystyle \lim_{x\to \infty}g(x)$ and $\displaystyle\lim_{x\to -\infty}g(x)$ both exist and are finite. Then $g$ is uniformly continuous.

Apply this to $g(x):=\int_{-\infty}^x f(t)dt. $
A: The compact support is irrelevant: as long as $f$ is measurable, bounded and Lebesgue integrable (actually, you could relax the latter to $1_{(-\infty,0]}f$ being integrable), the function $g$ will be Lipschitz continuous because for $s<t$ $$\lvert g(t)-g(s)\rvert\le\int_s^t\left\lvert f(x)\right\rvert\,dx\le M\lvert t-s\rvert$$
A: Define $F(x)=\int_{-\infty}^{x}f(t)dt$. It is well-known that $F$
is absolutely continuous, which implies that $F$ is uniformly continuous.
Note that the only condition is "$f$ is Lebesgue integrable".
We do not require that $f$ is bounded nor having compact support.
Proof: That proposition is well-known and the proof can be found in
every real analysis textbook. For the sake of completeness, I include
it.
Claim 1: If $f$ is measurable and bounded, then for each $\varepsilon>0$,
there exists $\delta>0$ such that $\int_{A}|f|<\varepsilon$ whenever
$m(A)<\delta$.

Proof of Claim 1: Choose $M>0$ such that $|f(t)|\leq M$ for all
$t\in\mathbb{R}$. Let $\varepsilon>0$ be given. Choose $\delta=\frac{\varepsilon}{2M}.$
For any measurable set $A$ with $m(A)<\delta$, we have that $\int_{A}|f|\leq M\cdot m(A)<\varepsilon.$

Claim 2: If $f$ is integrable, then for each $\varepsilon>0$, there
exists $\delta>0$ such that $\int_{A}|f|<\varepsilon$ whenever $m(A)<\delta$.
Proof of Claim 2: For each $n\in\mathbb{N},$ define $f_{n}=\min(|f|,n)$,
then $|f_{n}|\leq|f|$ and $f_{n}\rightarrow|f|$. By Lebesgue Dominated
Convergence Theorem, we have that $\int\left|f_{n}-|f|\right|\rightarrow0$.
Let $\varepsilon>0$ be given. Choose $n$ such that $\int\left|f_{n}-|f|\right|<\varepsilon.$
Note that $f_{n}$ is bounded and integrable, so we may choose $\delta>0$
such that $\int_{A}|f_{n}|<\frac{\varepsilon}{2}$ whenever $m(A)<\delta$.
Now, let $A$ be an arbitrary measurable set with $m(A)<\delta$,
we have that
\begin{eqnarray*}
 &  & \int_{A}|f|\\
 & = & \int_{A}\left(|f|-f_{n}\right)+\int_{A}f_{n}\\
 & \leq & \int\left|f_{n}-|f|\right|+\int_{A}|f_{n}|\\
 & < & \varepsilon.
\end{eqnarray*}

Now, the uniform continuity of $F$ follows immediately. For, let
$\varepsilon>0$ be given. Choose $\delta>0$ such that $\int_{A}|f|<\varepsilon$
whenever $A$ is measurable and $m(A)<\delta$. For any $x_{1}<x_{2}$
with $x_{2}-x_{1}<\delta$, we have that
\begin{eqnarray*}
|F(x_{2})-F(x_{1})| & = & \left|\int_{(x_{1},x_{2}]}f\right|\\
 & \leq & \int_{(x_{1},x_{2}]}|f|\\
 & < & \varepsilon.
\end{eqnarray*}
