$f: \mathbb{R} \to \mathbb{R}$ integrable, $F(x) = \int_a^x f(y)\,dy$, $F$ necessarily continuous Suppose $f: \mathbb{R} \to \mathbb{R}$ is integrable, and we define$$F(x) = \int_a^x f(y)\,dy.$$Why does it follow that $F$ is necessarily a continuous function?
 A: One possible answer would be
$$ |\int_a^{x+h} fd\lambda - \int_a^x fd\lambda| \le \int |1_{ [x, x+h]}f| d\lambda$$
Now, since $1_{[x, x+h]}f$ converges to 0 almost everywhere on $\mathbb{R}$ for $h \to 0$ and $|1_{[x, x+h]}f| \le |f|$ integrable, it follows from the Lebesgue convergence theorem that the RHS converges to 0, which is just the continuity claim. 
A: Let's repeat BST's proof, but use dominated convergence only for sequences.  
Fix $c \ge a$.  Show $F$ is continuous at $c$.
To show:
$$
\lim_{x \to c} F(x) = F(c)
$$
Assume not.  Then there is a sequence $x_n \to c$
such that $F(x_n) \to F(c)$ fails.  [Either it does not converge, or converges to something other than $F(c)$.]  Now
$$
F(x_n) = \int_{[a,x_n]} f(y) \;dy
=\int f(y)\;\mathbf{1}_{[a,x_n]}(y)\;dy
$$
Also $f(y)\;\mathbf{1}_{[a,x_n]}(y) \to f(y)\;\mathbf{1}_{[a,c]}(y)$ for almost all $y$ [indeed, for all $y$ except possibly $y=c$].  And 
$$
\big|f(y)\;\mathbf{1}_{[a,x_n]}(y)\big| \le \big|f(y)\big|.
$$
Thus, by the dominated convergence theorem, $F(x_n) \to F(c)$.  This contradiction completes the proof.
