If $f(x)$ is measurable function for $x\in\mathbb{R}$, then if $h(x,t)=f(x+t)$ also measurable in $<x,t>\in\mathbb{R}^2$

Question

So for $h(x,t)$ to be measurable it means for any $a\in\mathbb{R}$ that $Z(a)=\left\{<x,t>\big|h(x,t)\geq a\right\}$ is a measurable set

Since $f(x)$ is measurable, then we have $E(a)=\left\{y\big|f(y)\geq a\right\}$ is measurable.

Thus I need to prove that $F(a)=\left\{<x,t>\big|x+t\in E(a)\right\}$ is a measurable set in $\mathbb{R}^2$

Now I can assume that $h(x,t)=x+t$ which is obviously an measurable function and trying to prove that the pre-image of $E(a)$ in $h(x,t)$ is measurable.

In other words, if the pre-image of $h(x,t)$ for a measurable set is also measurable??

Based on what I know, it is not true that the pre-image of a measurable set through a measurable function is also a measurable set.

Thus I'm here asking if it is true. And if you could be so kind to provide the prove. Thanks a lot!

Answer 1

Actually, the measurability of $h$ follows from :

1. the measurability of $(x, t) \in \mathbb{R}^2 \mapsto x + t$ (it is even continuous) and of $f$
2. the fact that composition preserves measurability.

Using property 2 above can help a lot when proving something is measurable (and this way you avoid considering explicit sets). Other interesting stability properties of measurable functions are stated here.