Proving f = 0 almost everywhere

[A.Magnus]

I am working on a problem$^{(1)}$ in Measure & Integration (chapter on Product Measures) like this:

Suppose that $f$ is real-valued and integrable with respect to 2-dimensional Lebesgue measure on $[0, 1]^2$ and also

$\int_{0}^{a} \int_{0}^{b} f(x, y) dy dx = 0$
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for all $a, b \in [0, 1].$ Show that $f = 0$ almost everywhere.
​

Here are what I only know of this problem:

(1) The mapping is $f : [0, 1] \times [0, 1] \longrightarrow \Bbb R$

(2) Since we need to prove $f = 0$ a.e., and borrowing from set builder of convergence a.e., I think I need to prove this:

$\mu (\{ (x, y) \in [0, 1]^2 \mid f(x, y) \neq 0\}) = 0.$

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Since this problem is under chapter on Product Measures involving iterated integral, I suspect I need to use the Fubini Theorem, but I do not know how to do it. Any hints or helps would be very much appreciated. Thank you very much for your time and effort.______________________________________________________________________________

(1) Richard F. Bass' http://homepages.uconn.edu/~rib02005/real-analysis-04nov2011.pdf 2nd. edition, chapter 11: Product Measure, Exercise 11.6, page 88.

 

[RUber]

Assume that f is non-zero on a set that is not measure zero. Can you have the integral be zero? (yes) However, you can partition that set into the set where f is positive and the set where f is negative. It is impossible for the integral to be zero over both sets.
Apply that same rationale to your argument with the limits of integration a and b.

 

[A.Magnus]

Assume that f is non-zero on a set that is not measure zero. Can you have the integral be zero? (yes) However, you can partition that set into the set where f is positive and the set where f is negative. It is impossible for the integral to be zero over both sets.
Apply that same rationale to your argument with the limits of integration a and b.

Thank you to RUber for your speedy response. Allow me though to ask you one question:

Am I in the right direction on (2)? In other words, are you giving me the hints above within the framework of $\mu (\{(x, y) \in [0, 1]^2 \mid f(x, y) \neq 0 \}) = 0$?

Thank you again for your time and effort.

 

[RUber]

That's right. I propose proof by contradiction.

 

[WWGD]

To add a bit to RUber's idea, illustrating, assume you have a small square s * (with sides parallel to the x-axis an y-axis respectfully) with measure>0 ,where $f \neq 0$ , and so that $f == 1$ in the left half and $f == -1$ on the right half. Remember that for simple functions f with values $c_i$: ,$\int f = \int (\Sigma c_i \ chi_{c_i} ) = \Sigma \ c_i m( x: f(x)=c_i)$ . Now remember that the integral must be zero over _all_ partitions.

* Just to illustrate; obviously this set does not have to be a square; it doesn't even have to contain a square of measure >0..

 

[A.Magnus]

That's right. I propose proof by contradiction.

To add a bit to RUber's idea, illustrating, assume you have a small square s * (with sides parallel to the x-axis an y-axis respectfully) with measure>0 ,where $f \neq 0$ , and so that $f == 1$ in the left half and $f == -1$ on the right half. Remember that for simple functions f with values $c_i$: ,$\int f = \int (\Sigma c_i \ chi_{c_i} ) = \Sigma \ c_i m( x: f(x)=c_i)$ . Now remember that the integral must be zero over _all_ partitions.

* Just to illustrate; obviously this set does not have to be a square; it doesn't even have to contain a square of measure >0..

Here are what I managed to come up after feedbacks from you two, but I am still lost after the first two steps:

(1) By way of contradiction, assume that $f = 0$ a.e. is not true, meaning that $\mu (\{ (x, y) \in [0, 1]^2 \mid f(x, y) \neq 0\}) \neq 0.$
(2) Let $A = \{ (x, y) \in [0, 1]^2 \mid f(x, y) \neq 0\}$, that is $f(A) \neq 0$
(3) ...
(4) ...

Any further hints would be very much appreciated. Thank you for your time.

 

[WWGD]

Basically, look for a rectangle $a \times b$ , for some specific values of $a,b$ that will contain a subset of the set of measure non-zero where the positive part does not cancel out the negative part in the integral. You can use a rectangle for S and the characteristic function as a motivating example (ultimately using simple functions to approximate your measurable function may be necessary.) so that the rectangle $[0,a] \times [0,b]$ intersects S at S' in such a way that $\int_{S'} f \neq 0$ , following up on RUber's idea. Remember that the integral of f is assumed to be 0 over _all_ rectangles $[0,a] \times [0,b] ; 0<a,b<1$.

 

[Dick]

Basically, look for a rectangle $a \times b$ , for some specific values of $a,b$ that will contain a subset of the set of measure non-zero where the positive part does not cancel out the negative part in the integral. You can use a rectangle for S and the characteristic function as a motivating example (ultimately using simple functions to approximate your measurable function may be necessary.) so that the rectangle $[0,a] \times [0,b]$ intersects S at S' in such a way that $\int_{S'} f \neq 0$ , following up on RUber's idea. Remember that the integral of f is assumed to be 0 over _all_ rectangles $[0,a] \times [0,b] ; 0<a,b<1$.

Did you people, WWGD and RUber actually solve this problem? I think it's a bit harder than you are thinking. You (A.Magnus) should easily be able to show given the premises that the integral over ANY rectangle in the domain is zero. This isn't a Riemann integral, it's a Lebesgue integral. You could imagine that you would start it a function that's 1 over half the domain and -1 over half the domain and then somehow scramble it up so EVERY rectangle contains a equal amount of 1 and -1. That's what you need to show is impossible. I think you need something like http://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem or http://en.wikipedia.org/wiki/Lebesgue's_density_theorem.