Conditions a discontinuous f must obey so $\int f$ is continuous? What conditions must a discontinuous $f$ satisfy for the $\int\limits_{-\infty}^x f(u) \ dP$ to be continuous?
I would say the set of discontinuities must have measure zero. Is it also sufficient?
Am I missing something?
Any help would be appreciated.
 A: Let $f$ be integrable over each set $(-\infty,a)$, $a\in\mathbb R$. Then, if $x > x_0$,
$$
\int_{(-\infty,x]}f\,dP - \int_{(-\infty,x_0]}f\,dP = \int_{(x_0,x]}f\,dP = \int_{(-\infty,x_0+1)}\chi_{(x_0,x]}f\,dP.
$$
The integral on the RHS tends to zero as $x$ approaches $x_0$ by Lebesgue's theorem. Hence, the function is always right-continuous. If $x < x_0$, then
$$
\int_{(-\infty,x_0]}f\,dP - \int_{(-\infty,x]}f\,dP = \int_{(x,x_0]}f\,dP = f(x_0)P(\{x_0\}) + \int_{(-\infty,x_0+1)}\chi_{(x,x_0)}f\,dP.
$$
Hence, the limit as $x\to x_0$ is zero if and only if either $f(x_0) = 0$ or if $P$ does not have mass at $x_0$.
Here, $(x_0,x]$ means either $(x_0,x]$ when $x > x_0$ or $[x,x_0)$, when $x < x_0$.
The discontinuities of $f$ are not important here. For example, let $f=1$ and let $P$ be the probability measure that has mass one at zero. Then
$$
\int_{(-\infty,x]}f\,dP = \begin{cases}0 &\text{if }x < 0\\1 &\text{if }x\ge 0\end{cases}.
$$
So, there is a jump at zero although $f$ is continuous.
