Here's the statement of the problem:
Let $f \in L^1([0,1]^2,\mathscr{B}(\mathbb{[0,1]^2}),m)$, where $\mathscr{B}(\mathbb{[0,1]^2})$ is the Borel $\sigma$-algebra and $m$ is the two-dimensional Lebesgue measure. Prove that if $\int_{0}^{a} \int_{0}^{b}f(x,y)dydx = 0$ for all $a, b \in [0,1]$, then $f=0$ $m$-a.e.
So the idea of the proof is to let $\mathscr{C}=\{E\subset[0,1]^2: \int_{E}f dm=0\}$ and show that $\mathscr{C}$ contains all sets that are the product of two intervals and that it is a $\sigma$-algebra. Then it contains the Borel $\sigma$-algebra. This gives the desired result by Prop 8.1 in the Bass Real Analysis book.
My question is in showing $\mathscr{C}$ is a $\sigma$-algebra. I've done everything but show it is closed under countable unions... I've tried taking a collection of sets in $\mathscr{C}$ and breaking them into a disjoint collection with the same union but then can't confidently prove that this integral is zero.
Could someone help?
Your idea seems plausible, I suspect it needs the $\pi-\lambda$ Theorem.
But it seems that the reasoning goes through:
It is not hard to see that \begin{align*} \iint_{[a,b]\times[c,d]}f(x,y)dxdy=0, \end{align*} so the measure $\mu$ defined by \begin{align*} \mu([a,b]\times[c,d])=\iint_{[a,b]\times[c,d]}f(x,y)dxdy \end{align*} is actually the zero functional, and in fact it is the Radon-Nikodym of $f$, that is, \begin{align*} f=\dfrac{d\mu}{dm},~~~~\text{a.e.}, \end{align*} then $f=0$ a.e. since $0$ can be a Radon-Nikodym and each pair of two Radon-Nikodym differs up to a.e. only.