$f: \mathbb{R} \to \mathbb{R}$ integrable, $F(x) = \int_a^x f(y)\,dy$, $F$ necessarily continuous

Question

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?

Answer 1

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.

Answer 2

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.