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How to prove that the graph of measurable function from $R^d\to R$ is of measure zero?

I proved for a continuous function but for any measurable function is more subtle.

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The graph $\Gamma$ of the function is measurable. The vertical sections $\Gamma_x=\{y:(x,y)\in \Gamma\}$ are measurable too and contain a single point, a set of measure zero. By Fubini's theorem for null sets, $\Gamma$ has measure zero.

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The proof of measurability of the graph you link to seems a bit complicated: let $(I_n)_{n \in \mathbb{N}}$ be an enumeration of the non-trivial intervals with rational endpoints. Then $$f(x) = y \iff \forall n \left(y \in I_n \implies f(x) \in I_n\right)$$ shows that $\Gamma$ is measurable. – commenter Oct 23 '12 at 0:00
For those not as comfortable with the logical way of stating it, a translation to more set theoretic language would be $\Gamma = \cap_n( (\mathbb{R}^d \times I_n^c) \cup (f^{-1}(I_n) \times I_n))$ – nullUser Jul 27 '15 at 22:09

Here is a more elementary (and far longer) solution in case someone looks and doesn't know Fubini's theorem (because I just looked and don't know it).

let $f : \mathbb{R}^d \rightarrow \mathbb{R}$ be a measurable function. Then let $$F_n = \min(f(x), n)\chi_{[-n,n]^d}$$ be a horizontally and vertically truncated version of $f$. Then we have $\Gamma(f) \subseteq \bigcup_n \Gamma(F_n)$, so it suffices to show that $\Gamma(F_n)$ is a null set for all $n$.

To do this, fix $N$ and partition $[-N,N]$ into $2mN$ disjoint intervals $I_j^m$ of size $\frac{1}{m}$. And let $A_j^m = F_N^{-1}(I_j^m)$ be the preimage of $I_j^m$ under $F_N$, which is measurable because $F_N$ is a measurable function.

Then $$\Gamma(F_N) = \bigcap_{m=1}^{\infty} \bigcup_{j=1}^{2Nm} A_j^m \times I_j^m,$$ so $\Gamma(F_N)$ is measurable and $\bigcup_{j=1}^{2Nm} A_j^m \times I_j^m$ is a decreasing sequence in $m$, all of which have finite measure, so by the downward monotone convergence of sets $$\mu(\Gamma(F_N)) = \lim_{m\rightarrow\infty} \mu(\bigcup_{j=1}^{2Nm} A_j^m \times I_j^m ) \leq \lim_{m\rightarrow\infty} \sum_1^{2Nm} \mu(A_j^m) \frac{1}{m}$$ where the inequality is by subaddivitiy and the fact that $\mu(A_j^m\times I_j^m) = \mu(A_j^m)\mu(I_j^m) = \mu(A_j^m)\frac{1}{m}$. Then by construction the $A_j^m$'s are disjoint and all contained in $[-N,N]^d$, so $$ \lim_{m\rightarrow\infty} \sum_1^{2Mn} \mu(A_j^m)\frac{1}{m} \leq \lim_{m\rightarrow\infty} 2^dN^d \frac{1}{m} = 0.$$

Thus we have shown that $\mu(\Gamma(F_N)) = 0$ and we're done.

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Isn't there a problem in the last inequality? Maybe I'm mistaken, but I think the $m$ stemming from the sum in the LHS will cancel the $\frac{1}{m}$ in the RHS. – shamisen Jul 1 '14 at 0:10

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