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...


  • $\begingroup$ I think $F$ should be right continuous. $\endgroup$ – 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 $t<x_0$, 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 $\{x_0\}$).

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.