Suppose $\lambda$ is Lebesgue measure on $\mathbb{R}$ and $f:\mathbb{R}\to\mathbb{R}$ is a Borel measurable function such that $$\int_\mathbb{R} |f|d\lambda<\infty.$$ Define $g :\mathbb{R}\to\mathbb{R}$ by $$g(x) =\int_{(−\infty, x)}f d\lambda.$$ Prove that $g$ is uniformly continuous on $\mathbb{R}$.
In the book is the following result:
Integrals on small sets are small: Suppose $(X, \mathcal{S}, \mu)$ is a measure space, $g : X \to [0, \infty]$ is $\mathcal S$-measurable, and $\int_\mathbb{R}g d\mu < \infty$. Then for every $\varepsilon > 0$, there exists $\delta > 0$ such that $$\int_B g d\mu < \varepsilon$$ for every set $B \in \mathcal S$ such that $\mu(B) < \delta$.
My attempt: Let $\varepsilon>0$, then by the previous result, exists $\delta>0$ such that for every $B\in\{B\subset\mathbb{R}\mid B\text{ is Lebesgue-measurable and }\lambda(B)<\delta\}$ $$\int_B |f|d\lambda<\varepsilon$$ We can consider a partition $\{E_n\}_{n\in\mathbb{N}}$ of $\mathbb{R}$ such that $E_n\in\{B\subset\mathbb{R}\mid B\text{ is Lebesgue-measurable and }\lambda(B)<\delta\}$ for all $n\in\mathbb{N}$ and such that each $E_n$ is an interval, so for all $x,y\in\mathbb{R}$ $(x<y)$ such that $x,y\in E_n$ for some $n\in\mathbb{N}$: $$\left|\int_{(x,y)}fd\lambda\right|=\left|\int_{(-\infty,x)}fd\lambda-\int_{(-\infty,y)}fd\lambda\right|=|g(x)-g(y)|$$ and $$\left|\int_{(x,y)}fd\lambda\right|\leq\int_{(x,y)}|f|d\lambda\leq\int_{E_n}|f|d\lambda<\varepsilon$$ So, for every $\varepsilon>0$, there exists $\delta>0$ such that for any $x,y\in \mathbb R$ such that $|x-y|<\delta$, we have that $|g(x)-g(y)|<\varepsilon$ which is equivalent to saying that $g$ is uniformly continuous. $\square$
Is what I did correct? If not, any correction or hint you can give me? Thank you very much in advance.
