# Null function and Lebesgue measure on $\mathbb{R}^N$

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 $$(0,T)\times\Omega$$)?

• 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. Commented May 8 at 14:03
• Can you detail a little? What are the sets for which you apply this? Commented May 8 at 15:00

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 $$A=\{(t, x): f(t,x)\neq 0\}$$.