Let $f: X\to \mathbb R$ be $\mathcal{A}-\mathcal{B}-$measurable. If $\mu$ is $\sigma$-finite, show that for the product measure $\mu \otimes \beta^{1}$, $(\mu \otimes \beta^{1})( \operatorname{graph}(f))=0$
It is clear since $\beta^{1}(f(\{x\})=0$ for any $x$ as a singleton, however, I need to use the definition above to substantiate it.
My idea:
Since $\mu$ is $\sigma-$finite, consider the cover $X \subset \bigcup\limits_{n \in \mathbb N}A_{n}$
Note that $(\mu \otimes \beta^{1})( \operatorname{graph}(f))=(\mu \otimes \beta^{1})(\{(x,f(x)): x \in X\})\leq (\mu \otimes \beta^{1})(\{(x,f(x)): x \in \bigcup\limits_{n \in \mathbb N}A_{n}\})\leq \sum\limits_{n \in \mathbb N}(\mu \otimes \beta^{1})(\{(x,f(x)): x \in A_{n}\})$
Now I want to consider the interval $(x, \frac{f(x)-\epsilon}{2^{n}\mu(A_{n})}), (x, \frac{f(x)+\epsilon}{2^{n}\mu(A_{n})})$ where $\epsilon >0$ is arbitrary.
This leads to: $\sum\limits_{n \in \mathbb N}(\mu \otimes \beta^{1})(\{(x,f(x)): x \in A_{n}\})\leq\sum\limits_{n\in \mathbb N}\mu(A_{n})*\frac{2\epsilon}{2^{n}\mu(A_{n})}=2\epsilon\xrightarrow{}0$
Does this suffice?