If $f :\mathbb{R}\to\mathbb{R}$ is measurable, then $E = \{x: f(x) \geq 3\}$ is measurable

Question

Prove: Suppose $f : \mathbb{R}\to\mathbb{R}$ where $f$ is measurable and $E = \{x: f(x) \geq 3\}$. Show $E$ is measurable.

I saw this statement while reading in a paper and thought this might be a good proof to know. It was about Lebesgue measures which I am sort of familiar with. I did not learn this sort of proof though.

Can someone please show me how this proof would look like?

I came up with this:

$$E=f^{-1}([3,\infty))$$ Is this correct? Is it possible to break this down further? I am not sure since I did not learn anything about this. I am just trying this for fun. I would still like to see a proof of it.

Answer 1

Note that $$E=\{x\mid f(x)\ge 3\}=f^{-1}([3,\infty))$$ where $f^{-1}$ denotes the pre-image. What is the definition of a measurable function?