1
$\begingroup$

We know that some measurable function $f:(0,T)\times\Omega\to\mathbb{R}$, where $\Omega\subset\mathbb{R}^N$ has the following property:

For almost all $t\in (0,T)$ (Lebesgue measure in $\mathbb{R}$) we have that:

$f(t,x)=0$ for almost all $x\in\Omega$ (Lebesgue measure in $\mathbb{R}^N$).

Is it true that $f(t,x)=0$ for almost all (Lebesgue measure on $(0,T)\times\Omega$)?

$\endgroup$
2
  • 1
    $\begingroup$ This follows directly from the definition of product measures, as well as the fact that the $n$-dimensional Lebesgue measure is a product measure of Lebesgue measures. $\endgroup$ Commented May 8 at 14:03
  • $\begingroup$ Can you detail a little? What are the sets for which you apply this? $\endgroup$
    – Bogdan
    Commented May 8 at 15:00

1 Answer 1

3
$\begingroup$

This is an application of Tonelli's theorem. Namely, suppose $(X, \mu)$ and $(Y, \nu)$ are $\sigma$-finite measure spaces and $A\subset X\times Y$ is a measurable subset with the property that for $\mu$-a.e. $x$ the section $A_x:=\{y: (x, y)\in A\}$ has $\nu$-measure $0$. Then $$(\mu\times \nu)(A)=\int_{X \times Y} {1}_A d(\mu\times \nu)=\int_X \Big(\int_Y 1_{A_x} d\nu \Big) d\mu(x)=\int_X \nu(A_x) d\mu(x).$$ Here the integrand is $0$ almost everywhere hence the integral is $0$. That is, $(\mu\times \nu)(A)=0$. Now apply this to $A=\{(t, x): f(t,x)\neq 0\}$.

$\endgroup$
1
  • 1
    $\begingroup$ Very nice solution! Thank you very much! $\endgroup$
    – Bogdan
    Commented May 8 at 15:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .