Let $f:[0,a]\times[0,b]\rightarrow\mathbb{R}$ be a continuous function. Prove a version of Fubini's Theorem:

Question

$$\int_0^b\left(\int_0^af(x,y)dx\right)dy=\int_0^a\left(\int_0^bf(x,y)dy\right)dx.$$ I am given the following hint: Investigate $p:[0,b]\rightarrow\mathbb{R}$ defined by $$p(t)=\int_0^t\left(\int_0^af(x,y)dx\right)dy-\int_0^a\left(\int_0^tf(x,y)dy\right)dx.$$ I have read the classical proof of Fubini's Theorem that uses $\sigma$-finite spaces, but this is beyond the scope of which this question is asked in. I am looking for advice on how to utilize the provided hint to establish a proof of this.

Answer 1

$p'(t)=0$, where $\dfrac{d}{dt}\displaystyle\int_{0}^{a}\left(\int_{0}^{t}f(x,y)dy\right)dx=\int_{0}^{a}\left(\dfrac{d}{dt}\int_{0}^{t}f(x,y)dy\right)dx$ can be argued by Lebesgue Dominated Convergence Theorem.

Another method:

Let $\varphi(t)=\displaystyle\int_{0}^{a}\left(\int_{0}^{t}f(x,y)dy\right)dx$, then for small $h>0$ and $t<b$, \begin{align*} \left|\dfrac{1}{h}[\varphi(t+h)-\varphi(t)]-\int_{0}^{a}f(x,t)dx\right|&\leq\int_{0}^{a}\left(\dfrac{1}{h}\int_{t}^{t+h}|f(x,y)-f(x,t)|dy\right)dx\\ &=\int_{0}^{a}|f(x,\eta_{t,h})-f(x,t)|dx, \end{align*} where $\eta_{t,h}\in[t,t+h]$, now $f$ is uniformly continuous on $[0,a]\times[0,b]$, this may control the last integral going to zero.

Answer 2

If you know the Stone Weierstrass theorem: Verify that both iterated integrals give the same result for $f(x,y) = x^my^n, m,n \in \{0,1,2\dots \}.$ Thus both iterated integrals give the same answer for any polynomial in $x,y.$ Stone-Weierstass implies polynomials in $x,y$ are dense in $C([a,b]\times [c,d]).$ The result follows.