If $f(x)$ is measurable then $g(x,y)=f(x)-y$ is measurable

Question

Let $(X,\mathfrak A)$ be a measurable space and $f:(X,\mathfrak A)\to(\mathbb R_+,\mathfrak B^1)$ measurable, $f\geq 0$. Then the function $$g:X\times\mathbb R\to\overline{\mathbb R}\\ g(x,y) = f(x)-y$$ is $\;\mathfrak A\otimes\mathfrak B^1 - \overline{\mathfrak B} \;$ measurable.

This is how far I got: Measurability of $g$ means that $\{g>a\}\in\mathfrak A\otimes\mathfrak B^1$ for every $a$. In more detail: \begin{align}\{(x,y):g(x,y)>a\}&\in\mathfrak A\otimes\mathfrak B^1 \\ \Leftrightarrow\; \{(x,y):f(x)>a+y\}&\in \mathfrak A\otimes\mathfrak B^1 \end{align} Since $f$ is measurable, one has that the sets $\{x:f(x)>a+y\}\in \mathfrak A$ for all $a,y\geq 0$, but this doesn't really tell me anything about the measurability of the sets $$\{g>a\} = \bigcup_{y>0}\{x:f(x)>a+y\}\times\{y\} $$ Any help would be great...

Answer 1

Hint:

1. Show that the mappings $$(x,y) \mapsto f(x)$$ and $$(x,y) \mapsto -y$$ are measurable (you can prove this using the measurability criterium mentioned in your question).
2. Use that the sum of two measurable functions is again measurable.