Prove $F(x) = \int_{-\infty}^x f(t) dt$ is uniformly continuous.

Question

Following this question:$f$ is integrable, prove $F(x) = \int_{-\infty}^x f(t) dt$ is uniformly continuous.

Our question is that let $f\in L_1(\mathbb{R})$. Prove $F(x) = \int_{-\infty}^x f(t) dt$ is uniformly continuous.

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I have the following Lemma:

If for every $\epsilon>0$, there exists $\delta>0$ such that $\mu(A)=\mu([\min\{x, y\}, \max\{x,y\}])<\delta$, then $$ \int_{A}|f|<\epsilon. $$

Can we use this Lemma to prove our result? Since for every $\epsilon>0$, there exists $\delta>0$ so that $$ |F(x)-F(y)|\le \int_{A}|f|<\epsilon $$ whenever $\mu(A)<\delta$ and $A=[\min\{x, y\}, \max\{x,y\}]$.

Answer 1

Yes. Here, $\mu$ is Lebesgue measure, so $\mu (\{min \{x,y\}, \max \{x,y\})=|y-x|$ and the Lemma can be directly applied.