Prove graph of measurable function is measurable in $\sigma$-finite case and that the product measure is $0$.

Let $$(X, \mathcal{A}, \mu)$$ be a $$\sigma$$-finite measure space, and let $$f: X \to \mathbb{R}$$ be measurable. Then, $$\Gamma(f)$$, the graph of $$f$$ defined as

$$\Gamma = \{(x,y) \in X \times \mathbb{R}: f(x) = y\}$$

is measurable in the $$\sigma$$-algebra $$\mathcal{A \times L}$$ where $$(\mathbb{R},\mathcal{L},m)$$ is the measure space composed of the Lebesgue $$\sigma$$-algebra ($$\mathcal{L}$$) on $$\mathbb{R}$$ and the Lebesgue measure $$m$$.

Furthermore, prove that the product measure is $$0$$.

For the first part, I am trying to find the measurable rectangle to prove this is measurable in the product sigma algebra.

I know that $$\Gamma = X \times \{f(x)\}$$

$$X \in \mathcal{A}$$ trivially. Also, is the reason why $$\{f(x)\} \in \mathcal{L}$$ the fact that $$f$$ is measurable? I know that $$f$$ being measurable means that

$$\{x:f(x) > a\} \in \mathcal{A} \ \ \forall a \in \mathbb{R}$$

How does this translate to $$\{f(x)\}$$ being measurable on $$\mathcal{L}$$?

Furthermore, assuming this is proven. Let $$\chi_A$$ be the indicator function of some set $$A$$.

We have that the measure of $$\Gamma$$, by definition is

$$(\mu \times m) (\Gamma)=\int_\Gamma \mathrm{d}(\mu \times m) = \int_{X\times\mathbb{R}} \chi_\Gamma ((x,y)) \mathrm{d}(\mu \times m)$$

and since the indicator function is, by definition, non-negative, we can use Fubini's theorem to get

$$(\mu \times m)(\Gamma)=\int_X\int_\mathbb{R} \chi_{\{(x,y):f(x)=y\}} ((x,y)) \mathrm{d}m \mathrm{d}\mu$$ But here I have no idea on how to do the first integral or how to proceed in general from here.

Thank you so much!

• Unless $f$ is constant, it is not true that $\Gamma = X \times \{f(x)\}$. Oct 25, 2019 at 7:40

Hints: compose the maps $$(x,y) \to x-y$$ and $$(x,y) \to (f(x),y)$$. Verify that these two maps are measurable . Hence $$(x,y) \to f(x)-y$$ is measurable. The graph of $$f$$ is just the inverse image of $$\{0\}$$ under this map.

For the second question just note that for any given $$x$$ there is only one $$y$$ such that $$f(x)=y$$. Hence when you integrate w.r.t. $$y$$ first you bet $$m(\{f(x)\}$$, Since Lebesgue measure of any singleton set is $$0$$ we get $$(\mu \times m)(\Gamma)=\int 0 d\mu=0$$.

• Why is $\Gamma_f$ measurable if the inverse image of $\{0\}$ is $\Gamma_f$? Oct 25, 2019 at 10:10
• Because $\{0\}$ is a Borel set in $\mathbb R$ and $\Gamma_f$ is the inverse image of this Borel set under a measurable function. Oct 25, 2019 at 10:12
• Oh ok! And this is because the Borel $\sigma$-algebra is contained in the Lebesgue $\sigma$-algebra right? Oct 25, 2019 at 10:13