$f$ continuous $\Rightarrow$ $f$ $\mu$ measurable?

Question

Let $f: X\to [0,\infty]$ be a continuous function and $\mu$ an outer measure on X. For a continuous function I can split it in a sum of step functions, so $$ f(x) = \sum\limits_{i=1}^\infty s_i \chi_{A_i} $$ Is $f$ also $\mu$ measurable and therefore Lebesgue integrable?

If not can you give me an example? I am not even sure if the splitting is possible for every continuous function.

Edit:
I see that splitting isn't possible for every continuous function. After reading a bit in my notes of my lecture, I cant see why I can argument that a function $f$ is $\mu$ measurable if it is continuous. Has someone an explaining or a hint so I can try the prove myself?

Answer 1

Welcome to MSE!

We need a little bit of bonus information -- $X$ must be a topological space to make sense of continuous functions, and $X$ must also be equipped with a $\sigma$-algebra to make sense of measurable functions.

In order for continuity to imply measurability, these two structures must be compatible (in which case $\mu$ is called a borel measure). Without this constraint, things can go wrong. For instance if $X$ is equipped with the discrete topology, then every function is continuous. As an exercise, you might try to use this idea to find an explicit example of a space $X$ equipped with a topology and a $\sigma$-algebra so that some continuous function is not measurable.

Ok, so with the pedantic caveat out of the way, let's try to see why (for borel measures) a continuous function is automatically measurable. Here's a hint:

Let's look at the definitions of measurable and continuous functions.

• $f$ is continuous whenever $f^{-1}((a,b)) \subseteq X$ is open for every $(a,b) \subseteq \mathbb{R}$.
• $f$ is measurable whenever $f^{-1}((a,b)) \subseteq X$ is measurable for every $(a,b) \subseteq \mathbb{R}$.

So to show continuous $\Rightarrow$ measurable, it suffices to show that every open subset of $X$ is measurable (do you see why?). Can you use our assumption that $\mu$ is a borel measure to finish up?

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I hope this helps ^_^