How to show that $f=0$ a.e. on $[0,1]\times [0,1]$. I have question I'm not able to solve. I wouldn't mind full solutions as this isn't homework.  
Suppose $f(x,y)$ is a bounded and Lebesgue measurable function on $[0,1]\times [0,1]$. How to show that if $$ \int_a^b \int_c^d f(x,y)~dxdy =0$$ for all $0\le a\lt b \leq 1$ and $0\leq c \lt d\leq 0$, then $f=0$ a.e. on $[0,1]\times [0,1]$. 
 A: Ben Derrett's argument works as well as this one, but this is a bit shorter and I like to differentiate integrals.  Since $f$ is bounded, it is in $L^1_{loc}$.  Let $(x,y)$ be a Lebesgue point of $f$ and consider that 
$\frac{1}{4r^2} \int_{x-r}^{x+r} \int_{y-r}^{y+r} f(x,y) dydx = 0$
for all $r$.  Since these squares are nicely shrinking sets (in that the area of a square is comparable to the area of a circle) Lebesgue's differentiation theorem gives that $f(x,y)$ = 0.  Since Lebesgue points are a full measure set $f = 0$ a.e.
Edit:
Lebesgue's differentiation theorem on $\mathbb{R}^n$ says that if $B_r(x)$ is the ball of radius $r$ around $x$ and $f$ is an $L^1_{loc}$ function, then for almost every $x$ we have $f(x)=\lim_{r\ \to 0} \frac{1}{m(B_r(x))}\int_{B_r(x)}f dm$ where m is the Lebesgue measure.  An easy lemma shows that if a sequence of sets shrinks to zero at the same rate as balls (they are called 'nicely shrinking sets') then the differentiation theorem still holds.
A: Let's define measures $\mu^+$ and $\mu^-$ on Lebesgue-measurable subsets of $[0,1]\times [0,1]$ by
$$\mu^+(E)=\int_E f^+(x,y)dxdy$$
and
$$\mu^-(E)=\int_E f^-(x,y)dxdy,$$
where $f^+$ and $f^-$ are the positive and negative parts of $f$.
It's a simple exercise (check!), using boundedness of $f$ (a condition we could relax), that these are bona fide measures. Consider the family of boxes
$$\mathcal{A}=\{[a,b]\times[c,d]:0\leq a\leq b \leq 1,~ 0\leq c\leq d\leq 1\}.$$
This is a $\pi$-system, since the overlap of two boxes is again a box. $\mu^+$ and $\mu^-$ agree on this $\pi$-system, and hence (see these notes or try yourself using Dynkin's Lemma) on the $\sigma$-algebra generated by the $\pi$-system. So they agree on $\mathcal{B}([0,1]\times[0,1])$. Both measures are $0$ on sets of measure $0$, so in fact they agree on the completion of $\mathcal{B}([0,1]\times[0,1])$, $\text{LEB}([0,1]\times[0,1])$.
So if $E\in \text{LEB}([0,1]\times[0,1])$,
$$\mu^+(E)-\mu^-(E)=\int_E f(x,y)dxdy=0.$$
Since $f$ is Lebesgue-measurable, we may consider the set on which $f$ is positive. The integral over this set is $0$, so, for any $n\in\mathbb{N}$, $f$ is less than $1/n$ a.e. (why?), so $f$ is non-positive a.e.. Then we may similarly show that it's non-negative, and thus $0$ a.e..
