Show that $g(x)=\int_{-\infty}^x f(t) \ dt$ is continuous.
Solution 1:
We have $$|g(x+h) - g(x)| \leq \left| \int_x^{x+h}f(t) dt \right| = \left| \int 1_{[x,x+h]}(t) f(t) dt \right|$$
As $h \to 0$, we have that $1_{[x,x+h]}(t) f(t)$ converges pointwise to $1_{\{x\}}(t) f(t)$. By the dominated convergence theorem we get: $$\left| \int 1_{[x,x+h]}(t) f(t) dt \right| \to \int 1_{\{x\}} f(t) dt = 0 \qquad \text{ as } h \to 0$$
Showing that $g$ is continuous.