TITLE: Is an integral always continuous?
QUESTION [42 upvotes]: Say I have a function $f(x)$ on some interval $[a,b]$. 
Say it is integrable such that $\displaystyle\int f(x)~dx $ is defined.
Is $\displaystyle\int f(x)~dx $ necessarily continuous? 
If I were to know that the integral is integrable itself such that $\displaystyle\int \int f(x)~dx $ is defined. Would that change anything? 
If so why?
Thank you

REPLY [5 votes]: I would like to add something to Pedro Tamaroff's answer. This would hold not only for Riemann-integrable functions but even Lebesgue-integrable ones. 
Let $(X,\mu)$ be any measure space and $f$ be $\mu$-integrable. One can approximate $f$ (in general $\mathscr{L}^p$ functions) by simple functions. So given $\epsilon>0$ we can find a simple function $\phi$ such that $\int|f-\phi|d\mu<\frac{\epsilon}{2}$. Let $|\phi(x)|< M$ for all $x$. Now for any measurable set $A$ we have $$|\int\limits_Afd\mu-\int
\limits_A\phi d\mu|\leq\int\limits_A|f-\phi|d\mu<\frac{\epsilon}{2}$$
or $$|\int\limits_Afd\mu|<\frac{\epsilon}{2}+|\int\limits_A\phi d\mu|\leq\frac{\epsilon}{2}+M\mu(A)<\epsilon$$ whenever $\mu(A)<\frac{\epsilon}{2M}=\delta$.
Now for a Lebesgue-integrable function $f$ over $[a,b]$ we define $F(x)=\int\limits_a^x fd\lambda$. Then given $\epsilon>0$ we can find (as above) some $\delta>0$ such that $|F(x)-F(y)|=|\int\limits_y^xfd\lambda|<\epsilon$ whenever $|x-y|<\delta$. This proves the (uniform) continuity of $F$.
Moreover, this function is differentiable almost everywhere.