# proving a function defined as an integral is continuous

Given $$f:(\mathbb{R},\mathbb{B}(\mathbb{R})) \rightarrow (\mathbb{R}, \mathbb{B}(\mathbb{R}))$$ is a non-negative function, such that $$\int_{\mathbb{R}}fd\mu < \infty$$. Given $$F:\mathbb{R} \rightarrow \mathbb{R}, F(x):= \int_{(-\infty, x)}f d\mu, \forall x \in \mathbb{R}$$. Show that $$F$$ is continuous.

Notation used: $$\mu$$ denotes the lebsgue measure and $$\mathbb{B}(\mathbb{R})$$ denotes the Borel sigma-algebra.

What I've tried:

My initial thoughts were to attempt to use sequential continuity to prove this. We can define an increasing sequence of simple functions $$(\phi_{n})_{n \in \mathbb{N}}$$ such that $$\lim_{n \to \infty} \phi_{n}(x) = f(x).$$

My other thought was to try and use the Dominated Convergence Theorem (DCT), since we have a sequence of simple functions as above.

My issues with using DCT are:

1. I'm not sure what to take as the dominating function;
2. In DCT, it is the limit of the sequence of functions that we obtain as the result. But we already know what the sequence of simple functions limit is...

Thanks.

• I think $F$ should be right continuous. – Amirhossein Aug 20 '19 at 14:09

Your idea using DCT works. To show that $$F$$ is continuous in any $$x_0\in\mathbb{R}$$, consider a sequence $$(x_n)$$ with $$x_n\rightarrow x_0$$. Then the sequence of functions $$h_n(t):=f(t)\,\chi_{(-\infty,x_n)}(t)$$ converges pointwise a.e. to $$h(t):=f(t)\,\chi_{(-\infty,x_0)}(t)$$, and $$|h_n(t)|\leq f(t)$$. Since $$f$$ is integrable, DCT implies $$\lim_{n\rightarrow\infty}F(x_n) = \lim_{n\rightarrow\infty}\int_{\mathbb{R}} h_n(t)\, dt = \int_{\mathbb{R}} h(t)\, dt = F(x_0).$$
Verification that $$h_n$$ converges pointwise a.e. to $$h$$: Fix $$t\in\mathbb{R}$$. If $$t>x_0$$, then for $$N$$ sufficiently large we have $$t>x_n$$ for all $$n>N$$, and consequently $$h_n(t)=0$$ for all $$n>N$$. In this case we therefore have $$\lim_{n\rightarrow \infty}h_n(t)=0=h(t)$$. If $$t, then for $$N$$ sufficiently large $$t for all $$n>N$$, and consequently $$h_n(t)=f(t)$$ for all $$n>N$$. In this case we therefore also have $$\lim_{n\rightarrow \infty}h_n(t)=f(t)=h(t)$$. We have thus established convergence of $$h_n(t)$$ to $$h(t)$$ for all $$t\in\mathbb{R}\setminus\{x_0\}$$. This means that $$h_n$$ converges pointwise to $$h$$ almost everywhere (the only exception being the null set $$\{x_0\}$$).
• Thanks, I just had a couple of follow up questions. How would one formally show that $h_{n}(t)$ converges pointwise a.e. to $h(t)$. In the final calculation, how do we know that we can bring the limit inside the integral? Finally, in the final calculation the domain on integration is over $\mathbb{R}$, when I wrote out the calculation by hand I got the following, I just wanted to check it's equivalent... $\lim_{n \to \infty} F(x_{n}) = \lim_{n \to \infty} \int_{(-\infty}, x_{n})}fd\mu = \lim_{n \to \infty} \int_{(- \infty, x_{n})}h_{n}(t)dt = \int_{(- \infty, x_{0})}h(t)dt = F(x_{0})$ Thanks. – VBACODER Aug 22 '19 at 8:41