Prove that $\int_a^b[\int_Xf(x,t)d\mu(x)]dt=\int_X[\int_a^b f(x,t)dt]d\mu(x)$

Question

Let $X$ be a measurable space and $f:X\times[a,b]\to\mathbb R$ a function such that for all $t\in[a,b],f(\cdot,t)$ is measurable. If for all $x\in X, f(x,\cdot)$ is continuous over $[a,b]$ and if there exist a function Lebesgue integral over $X$ such that $\vert f(x,t)\vert\le g(x)$ for all $x\in X$ and $t\in[a,b]$ prove that $$\int_a^b[\int_Xf(x,t)d\mu(x)]dt=\int_X[\int_a^b f(x,t)dt]d\mu(x)$$

By the DCT, we can have that $f\in L(\mu)$.

Then I am not sure on how to proceed. I thought in using the definition $\displaystyle\int_X fd\mu=\int_X f^+d\mu-\int_X f^-d\mu$ and then the indicator function to change from $X$ to a subset of $X$, say $A$ where a is an interval.

Though I'm not sure about it.

Can someone please help me with this?

Thank you in advance

Answer 1

Surely it is $dt$ on LHS of integral. Other than this, since $|f(x,t)|\leq g(x) \in L(\mu)$, we can have that $\int_{a}^b \int_{X} |f(x,t)|d\mu(x) dt <\int_{a}^b \int_{X} g(x)d\mu(x) dt = (b-a)||g||_{L^1(X)}<\infty$, from here we can apply Fubini's theorem. We would require the measure space $X$ to be sigma-finite to do this.

Answer 2

George Dewhirst's proof has a gap: To apply Fubini theorem, one should check that $f$ is jointly measurable (i.e., measurable with respect to the product $\sigma$-algebra.). However, this fact is not given.

Please check: https://en.wikipedia.org/wiki/Fubini%27s_theorem

"Failure of Fubini's theorem for non-measurable functions"

If I have time, I will go back to this question.