Is an integral always continuous?
Mia Lopez 
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
$\endgroup$ 93 Answers
$\begingroup$One can prove the following
THM Let $f:[a,b]\to\Bbb R$ be Riemann integrable over its domain. Define a new function $F:[a,b]\to\Bbb R$ by $$F(x)=\int_a^x f(t)dt$$ Then $F$ is continuous. That is, the map $$f\mapsto \int_a^x f$$ sends $\mathscr R[a,b]$ to $\mathscr C[a,b]$.
PROOF Let $c\in[a,b]$. Then $$F(x)-F(c)=\int_c^x f(t)dt$$
Since $f$ is integrable, we know it is bounded, say $|f(x)|\leq M$ over $[a,b]$. Then $$ -\int_c^x M\;dt\leq \int_c^x f(t)dt\leq \int_c^xM \;dt$$
which means $$-M(x-c)\leq \int_c^x f(t)dt\leq M(x-c)$$
Thus we get $$|F(x)-F(c)|\leq M|x-c|$$
Taking $x\to c$ the squeeze theorem says $\lim\limits_{x\to c}F(x)=F(c)$. $\blacktriangle$
$\endgroup$ 6 $\begingroup$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.
$\endgroup$ $\begingroup$The following is worth pointing out:
Firstly, note that if we replace "continuous" with "differentiable", the new statement isn't true. For instance, the function $$\mathbb{R} \rightarrow \mathbb{R}$$ $$x \mapsto \begin{cases} 1, & x \geq 0 \\ -1 & x < 0\end{cases}$$ has $x \mapsto |x|$ as an indefinite integral, which is not differentiable.
We do however have the following: if $f$ is Riemann integrable over its domain and $F$ is defined as in Pedro's answer, then: $$f \mbox{ continuous at } x \rightarrow F \mbox{ differentiable at } x$$
Here's a nice article about such things. In the parlance of that article, we'd say that $|x|$ is an indefinite integral of the aforementioned step function, but not an antiderivative.
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