Does the graph of a measurable function always have zero measure?

Question: Let $(X,\mathscr{M},\mu)$ be a measure space and $f\colon X \to [0,+\infty[$ be a measurable function. If ${\rm gr}(f) \doteq \{ (x,f(x)) \mid x \in X \}$, then does it always satisfy $$(\mu \times \newcommand{\m}{\mathfrak{m}} \m)({\rm gr}(f)) = 0$$even if the domain space is not $\sigma$-finite? Here $\m$ denotes Lebesgue measure in $[0,+\infty[$.

Context: define the "shadows" $$\begin{align*} G_{f,<} &\doteq \{ (x,y) \in X \times \left[0,+\infty\right[ \mid y < f(x) \} \\ G_{f,\leq} &\doteq \{ (x,y) \in X \times \left[0,+\infty\right[ \mid y \leq f(x)\}.\end{align*}$$One can prove that $$(1) \qquad (\mu \times \m)(G_{f,<}) = \int_X f(x)\,{\rm d}\mu(x)$$ and that, if $\mu$ is $\sigma$-finite, that $$(2) \qquad(\mu \times \m)(G_{f,\leq}) = \int_X f(x)\,{\rm d}\mu(x).$$This implies that if $\mu$ is $\sigma$-finite, then $(\mu \times \m)({\rm gr}(f)) = 0$. But is this hypothesis really needed?

I have the following proofs, in case this helps anyone think:

• Proof of (1) without $\sigma$-finiteness: If $A \in \mathscr{M}$ and $f = \chi_A$, then $$(\mu \times \m)(G_{f,<}) \stackrel{(\ast)}{=} (\mu \times \m)(A \times [0,1[) = \mu(A)\m([0,1[) = \mu(A) = \int_X \chi_A(x)\,{\rm d}\mu(x),$$where $(\ast)$ holds because points have zero Lebesgue measure. The above clearly implies that the formula is valid for simple positive functions. If $f$ is measurable and positive, take a sequence $(\varphi_n)_n$ of simple and positive functions that converge to $f$, increasing. Then $\bigcup_n G_{\varphi_n,<} = G_{f,<}$, so upper continuity of the product measure and the Monotone Convergence Theorem give $$\begin{align*}(\mu \times \m)(G_{f,<})&=(\mu \times \m)\left(\bigcup_n G_{\varphi_n,<}\right) = \lim_n (\mu \times \m)(G_{\varphi_n,f})\\ &= \lim_n \int_X \varphi_n(x)\,{\rm d}\mu(x) = \int_X f(x)\,{\rm d}\mu(x),\end{align*}$$as wanted. This fails for $G_{f,\leq}$, since that set equality need not be true anymore. If we still had equality of measures, it would be fine, but I think that amounts to my initial question.

• Proofs assuming $\sigma$-finitess: apply Fubini-Tonelli as follows: $$\begin{align*} (\mu \times \m)(G_{f,<}) &= \int_{X \times [0,+\infty[} \chi_{G_{f,<}}(x,y)\,{\rm d}(\mu \times \m)(x,y) \\ &= \int_X \int_{[0,+\infty[} \chi_{[0,f(x)[}(y)\,{\rm d}\m(y)\,{\rm d}\mu(x) = \int_X f(x)\,{\rm d}\mu(x)\end{align*}$$Since points have zero Lebesgue measure, the same argument with $[0,f(x)]$ instead of $[0,f(x)[$ gives the formula for $G_{f,\leq}$. And using $\{f(x)\}$ instead along with $\m(\{f(x)\}) =0$ gives $(\mu \times \m)({\rm gr}(f))=0$.

Bonus track: does anyone know any results in this direction if we had another measure space as co-domain? Is it possible to have "fat" graphics?

Solution 1:

In the non-$\sigma$-finite case you have to be careful with which "product measure" you mean, because they are not all equivalent. For example, let $X = [0,1]$ with counting measure $\mu$, $Y = [0,1]$ with Lebesgue measure and $f(x) = x$. The "minimal" product measure on $X \times Y$ gives the graph of $f$ measure $0$, because $\int_Y \chi_{\text{gr}(f)}(x,y)\; dm(y) = 0$ for all $x$, but the "maximal" product measure gives it measure $1$, because $\int_X \chi_{\text{gr}(f)}(x,y)\; d\mu(x) = 1$ for all $y$.

Solution 2:

No. First, recall that the product measure is not unique for non-$\sigma$-finite spaces.

Let $(X,\mathscr{M},\mu)=(\mathbb{R},2^\mathbb{R},\#)$ and $f(x)=|x|$. We consider the product measure $$\nu(A)=\int_\mathbb{R}\#A^xdm(x)$$ where $A^x=\left\{y\in\mathbb{R}:(y,x)\in A\right\}$. Note that this is well-defined in the product $\sigma$-algebra of $X\times\mathbb{R}$ (for rectangles $A=E_1\times E_2$, where $E_1\subseteq\mathbb{R}$ and $E_2\subseteq\mathbb{R}$ is Borel, the map $x\mapsto\#A^x$ is simply $\#E_1\chi_{E_2}$).

It should be clear that $\nu(\operatorname{graph}(f))=\infty$.