This theorem comes from Munkres' Analysis on Manifolds:
Let $Q$ be a rectangle in $\mathbb{R}^{n}$; let $f:Q \to \mathbb{R}$; assume $F$ is integrable over $Q$.
a) If $f$ vanishes except on a set of measure zero, then $\int_{Q}f = 0$
b) If $f$ is non-negative and if $\int_{Q}f = 0$, then $f4 vanishes except on a set of measure zero.
What is it that these statements are trying to convey?
For part a) what I gather from me trying to piece things together is that, given a rectangle $Q$ we have a measure zero set, call it $B$, contained in $Q$, if it is not contained in $Q$, then this measure zero set will be contained in a union of rectangles and hence a "larger" rectangle. Is what is being said is that given a function $f$ defined over $Q$ (then in turn it would be defined on our set $B$) if the function vanishes everywhere except on $B$ then the "integral" of the function over $Q$ (which also includes $B$) would also be zero?
For part b) given the conditions, is what is being said that the function has some value other than $0$ over the smaller set $B \subset Q$?
Examining it more I'm seeing that these two statements are about the function specifically over the measure zero set. So it is saying the function is defined over the measure zero set and everywhere else it is not.
Is this the right interpretation?
EDIT: For clarification when we say the function "vanishes" we mean that it goes to zero correct? i.e $f(x) \to 0$? I ask because a counter example stressing the need for $\int_{Q}f = 0$ to exist using $[0,1]$ and $f(x) = 1\ \text{or}\ 0$ depending on $x$ being rational or irrational appears.
Best Answer
Firstly, when we say a function vanishes, we mean that it is equal to zero (not that it "goes to" zero).
Roughly speaking, a "measure zero" set, is supposed to mean "a very small set" For example, finite sets or countable subsets of $\Bbb{R}^n$ all have measure zero. They're small in the sense that they "have zero volume". So, these two theorems are trying to give a somewhat more precise meaning of in what sense a measure zero set is "small".
(a) If $f:Q \to \Bbb{R}$ is the zero function, then obviously $\int_Q f = 0$. What this theorem is telling you is that if instead you have an integrable function $f:Q \to \Bbb{R}$, such that the set of points $B = \{x \in Q|\, f(x) \neq 0\}$ has measure zero, then $\int_Q f = 0$. Loosely translated, it says that if the function is mostly zero, then the integral will also be zero.
(b) This is a partial converse to part (a).
In the manner you have phrased it, this is an incorrect statement. These theorems are trying to give an illustration of how "small/negligible" measure zero sets are. For your second sentence, no! The function $f$ is defined on all of $Q$... so I'm not sure what you intended to say.