Prove functions in L1 space are Uniformly Continuous

Question

Let $f \in L^1(R)$ and define $g(x)=\int_{(-\infty,x)} f d\mu$ where $\mu$ is Lebesgue measurable. Show that $g$ is uniformly continuous.

Answer 1

$f \in L^{1}(\mathbb{R})$ implies $|f(t)| \leq M$ a.e x.

Let $E = \{ t \in \mathbb{R} : |f(t)| \leq M\}$, then $\mu (\mathbb{R}\backslash E) = 0$.

Suppose $\epsilon > 0$, let $\delta < \frac{\epsilon}{M}$.

Then $|g(x) - g(y)| \leq \int_{ \mathbb{R}} |f(t)| \cdot \chi_{[x, y]}(t) d\mu = \int_{E} |f(t)| \chi_{[x, y]}(t)d\mu $

$\leq M \mu( E \cap [x, y]) \leq M |x - y| < \epsilon$ for all $x, y \in \mathbb{R}$ for which $|x -y| < \delta$.