I want to check if my attempt is valid for a problem.
Q. If $f : E \rightarrow R$ is L-measurable, prove that the graph of $f$ is a null set in $R^{n+1}$ ($\mu = $ Lebesgue measure)
Note: $E\subset R^n$, is a measurable set, hence $x\in R^n, y \in R, R^{n+1} = R^n \times R $
My Attempt:
By the corollary: $R^n = R^l \times R^m \ (n = l+m), $ Suppose X is a L-measurable subset of $R^l$ and Y is a L-measurable subset of $R^m$, then the cartesian product $X\times Y$ is an L-measurable subsest of $R^n$
Define the following measurable subsets
$A = \{ (x,y) \in R^{n+1} \ | \ 0 \leq y < f(x), \ x\in E \}$
$B = \{ (x,y) \in R^{n+1} \ | \ 0 \leq y \leq f(x), \ x\in E \}$
$C = \{ (x,y) \in R^{n+1} \ | \ -\infty \leq f(x) < 0, \ x\in E \}$
$D = \{ (x,y) \in R^{n+1} \ | \ -\infty \leq f(x) \leq 0, \ x\in E \}$
Then, letting $G=graph \ of \ f(x) =y $, we observe that
$\mu(G) = \mu(B-A) + \mu(D-C) $
After taking the x section of each sets A,B,C,D we see that by fubini's thm
$\mu(A) = \mu(B) = \int_{\{x\in E | 0 \leq f(x) \}}^{} f(x) dx$
$\mu(C) = \mu(D) = \int_{\{x\in E | f(x) < 0 \}}^{} f(x) dx$
Hence, $\mu(G) = 0$
Would appreciate any comments or more elegant/simple solutions!
Let $G_f$ denote the graph of $f$. If $\mu$ is Lebesgue measure on $\mathbb R^{n}$ and $\nu$ is Lebesgue measure on $\mathbb R$ then $(\mu \times \nu) (G_f)=\int_{G_f} d\mu \times \nu=0$ by Fubini/Tonelli Theorem because the section of $G_f$ by any point $x \in \mathbb R^{n}$ is the singleton set $\{f(x)\}$ and $\nu \{f(x)\}=0$ for all $x$.