# 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 $[0,1]\times [0,1]$.

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The integrals define a measure on a $\pi$-system generating $\mathcal{B}([0,1]\times [0,1])$. – Ben Derrett May 5 '12 at 15:36
@BenDerrett: I don't understand what you mean. – Kuku May 5 '12 at 15:42
Lebesgue density theorem – GEdgar May 5 '12 at 15:45

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.

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@kuku I have added an outline of what the differentiation theorem says. It is essentially the fundamental theorem of calculus for Lebesgue integrable functions. You can find a proof along with a careful statement (and a statement of the lemma I am using about nicely shrinking sets) in any graduate level analysis textbook that covers measure theory. The statement and terminology I am using comes from Rudin's Real and Complex Analysis. – Chris Janjigian May 5 '12 at 16:08

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..

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