TITLE: Measure theory showing integral of non-negative function is continuous
QUESTION [0 upvotes]: Let $f:(\mathbb{R}, B(\mathbb{R})\rightarrow(\mathbb{R},B(\mathbb{R}))$ be a non-negative function and $\int_{\mathbb{R}}f d \lambda < \infty$. $F:\mathbb{R} \rightarrow \mathbb{R}, F(x):= \int_{(- \infty, x)}f d \lambda , \forall x \in \mathbb{R}$
Where $d \lambda$ denotes the Lebesgue measure
Show that $F$ is continuous for all $x \in \mathbb{R}$
Attempt:
I want to argue by sequential continuity. Therefore I want to show $x_{n} \rightarrow x \Rightarrow F(x_{n}) \rightarrow F(x)$. Suppose $x_{n} \rightarrow x$ then $F(x_{n}) = \int_{(- \infty, x_{n})} f d \lambda$. But since $x_{n} \rightarrow x$ then $F(x_{n} \rightarrow F(x) = \int_{(- \infty, x)}f d \lambda$.
This seems incorrect and I am missing steps.

REPLY [0 votes]: Here is a well known theorem that implies your result: For every $\varepsilon > 0$, there is $\delta > 0$ such that $\lambda(A) \leq \delta \implies \int_{A}f\,d\lambda \leq \varepsilon$.
Proof: Let $\varepsilon > 0$. For measurable $A$ and $M > 0$,
$$\int_{A}f\,d\lambda = \int_{A}\min(f, M)\,d\lambda + \int_{A}(f - \min(f, M))\,d\lambda \leq M\lambda(A) + \int (f - \min(f, M))\,d\lambda.$$
Now by DCT, there is $M$ such that $\int (f - \min(f, M))\,d\lambda \leq \varepsilon$. So for $\delta = \varepsilon/M$,
$$\lambda(A) \leq \delta \implies \int_{A}f\,d\lambda \leq 2\varepsilon.$$
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