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!

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

1 Answer 1


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 $(\mu \times m)(\Gamma)=\int 0 d\mu=0$.

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

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