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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.

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  • $\begingroup$ I think $F$ should be right continuous. $\endgroup$ – Amirhossein Aug 20 '19 at 14:09
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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 $$ \lim_{n\rightarrow\infty}F(x_n) = \lim_{n\rightarrow\infty}\int_{\mathbb{R}} h_n(t)\, dt = \int_{\mathbb{R}} h(t)\, dt = F(x_0). $$

Addition:

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 $t<x_0$, then for $N$ sufficiently large $t<x_n$ 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 $\{x_0\}$).

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  • $\begingroup$ 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. $\endgroup$ – VBACODER Aug 22 '19 at 8:41
  • $\begingroup$ @VBACODER: I added a proof of the pointwise convergence in the answer. You can bring the limit inside the integral because of DCT (it is the main statement of DCT). I cannot check you calculation, something is wrong with the formatting. $\endgroup$ – StarBug Aug 22 '19 at 15:26
  • $\begingroup$ Thank you, that makes sense. $\endgroup$ – VBACODER Aug 22 '19 at 15:43

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