Why is $g(x)=\int_{(-\infty,x]}fd\mu$ is uniformly continuous in $\mathbb R$ when $f\in L^{1}(\mathbb R,\mu)$ and $\mu$ is de Lebesgue measure?

Question

Let $f\in L^{1}(\mathbb R,\mu)$ where $\mu$ is de Lebesgue Measure, how can I prove that $g(x)=\int_{(-\infty,x]}fd\mu$ is uniformly continuous in $\mathbb R$?

Answer 1

If not, there would be a $\varepsilon>0$, such that, for every $\delta>0$, there exist $x_\delta,y_\delta\in\mathbb R$, with $$ |x_\delta-y_\delta|<\delta \quad\text{and}\quad |g(x_\delta)-g(x_\delta)|\ge\varepsilon. $$ In particular, for $\delta=1/n$ we would have sequences $\{x_n\}$, $\{y_n\}$, such that $$ |x_n-y_n|<\frac{1}{n}\quad\text{and}\quad |g(x_n)-g(x_n)|\ge\varepsilon. $$

Case I. The sequence $\{x_n\}$ is bounded and hence it contains a convergent subsequence $x_{k_n}\to x^*$. Clearly $y_{k_n}\to x^*$ as well, and as $g$ is continuous, then $g(x_{k_n})\to g(x^*)$, $g(y_{k_n})\to g(x^*)$ and thus $|g(x_{k_n})-g(x_{k_n})|\to 0$. Contradiction.

Case II. The sequence $\{x_n\}$ is unbounded and hence it contains a subsequence $x_{k_n}\to \infty$ or $x_{k_n}\to -\infty$. Consider the first subcase (the second one is treated similarly). We can choose the sequence $\{x_{k_n}\}$ and $\{y_{k_n}\}$, so that $$ x_{k_1}<y_{k_1}<x_{k_2}<y_{k_2}<\cdots<x_{k_n}<y_{k_n}<\cdots $$ Then $$ \int_{\mathbb R}|f|\,dx\ge \sum_{j=1}^\infty \int_{x_{k_j}}^{x_{k_{j+1}}}|f|\,dx \ge \sum_{j=1}^\infty |g({x_{k_{j+1}}})-g({x_{k_j}})|\ge \sum_{j=1}^\infty\varepsilon=\infty. $$ Again leading to a contradiction.