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I am confusing about a subtle point in the following question

Let Q be a rectangle in $R^n$; let $f:Q-->R$ be a bounded function. Show that if $f$ vanishes except on a closed set $B$ of measure zero, then the integral of f over Q exists and equals zero

The phrase "f vanishes except on a closed set B of measure zero." Does it mean that f is defined to be non zero over the set B where B is a closed set and is of measure zero. Also, what happens if the adjective "closed" is left out, will the result still hold true. This question come from an exercises in Munkres' Analysis on Manifolds. Munkres never explicitly make any distinction when he says that a function vanishes except on a set D of measure zero, never specifying whether D is a closed set or not.

Thanks in advance.

dato datuashvili
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Seth
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1 Answers1

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That depends on the notion of integral. If you are talking about Riemann integral, then adjective close is very important: look at the function that is $0$ on irrational points and $1$ on rational. Then any upper Riemann sum is $1$, and any lower Riemann sum is $0$. Also in the statement it is not $f$ is defined to be non zero over the set $B$ more like $f$ is defined to be zero not on $B$ and anything on $B$.

If you are asking about Lebesgue integral than it is true for any set of measure zero, not necessarily closed.

P.S. The proof for closed measure zero set and Riemann integral is quite direct: by compactness argument you can cover it by the finite number of rectangles of arbitrary small volume. Note, that for Riemann integral to be defined one assumes that $f$ is bounded.

user68061
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  • Sorry, i am more confused now. The question saids that the function f vanishes except on a closed set B of measure zero. I am assuming that Munkres is using the Riemann integral. The function f which it vanishes is not necessarily closed, even though f is bounded and the set B which f do not vanishes on is a closed set. So how can i use a compactness argument? – Seth Jan 12 '14 at 11:05
  • B is a closed subset of a rectangle, thus compact. Now, B has measure zero, so it has a cover by rectangles of arbitrary small volume (less then $\varepsilon$). Now, you can choose a finite subcover of these cover by compactness. Then the riemann sum for any partition that contains these rectangles satisfies $M \varepsilon>S>-M \varepsilon$, where $|f| < M$. Well. that's it. – user68061 Jan 12 '14 at 12:22
  • i don't have to worry or do anything for the points that are on the boundary of B? – Seth Jan 12 '14 at 12:28
  • sure, because you can look for an open cover (for example, make the homothety of each rectangle in the cover by $1+\varepsilon$. Size of cover is still arbitrary small ) – user68061 Jan 12 '14 at 12:32
  • Basically the proof will be similar to the case where f just vanishes except on a set B of measure zero. The only minor differences is that B is closed. In both case where B is closed or not, B is still a subset of a rectangle and hence it is compact. So i can use it to my advantage. Is that correct? – Seth Jan 12 '14 at 12:48
  • compact sets are always closed. Not any subset of a rectangle is compact – user68061 Jan 12 '14 at 13:14
  • Sorry, maybe i misunderstood how you arrived at "B is a closed subset of a rectangle, thus compact". Compactness in R^n requires that it be closed and bounded. In the cases where if B is simply stated as a measure zero set without any qualification that it is a closed or not, i also don't understand how is conclude that it is compact. – Seth Jan 12 '14 at 14:30
  • let us continue this discussion in chat – user68061 Jan 12 '14 at 14:37
  • Sorry, I have some problems with chat, so I will continue here. If $B$ is closed then it is compact, because it is a subset of a rectangle thus bounded. If $B$ is just measure zero then, as I said earlier, the fact is wrong – user68061 Jan 12 '14 at 14:40
  • On page 93 section 11 of munkres, theorem 11.2, it states

    Let Q be a rectangle in R^n; let f:Q-->R be a bounded function. Let D be the set of points of Q at which f fails to be continouus. then the integral of f over Q exists if and only if D has measure zero in R^n.

    In the proof of the "if" part of the theorem, somewhere in the proof, it just assumes that Q is compact.

    The book can be accessed through here.

    http://fourier.math.uoc.gr/~papadim/calculus_on_manifolds/Munkres.pdf

     – Seth Jan 12 '14 at 16:31
  • The function that is zero outside the set of measure zero is not always continuous outside of this set. In Munkres the compactness of Q i.e. of a rectangle is used. Well, rectangle is closed and bounded... – user68061 Jan 12 '14 at 16:41
  • So what you are saying is that if the domain of a function is a closed rectangle, so any subset of a closed rectangle is closed and bounded, hence compact. If that is the case, why make a distinction between in one case in the theorem where a set is of measure zero without any qualification whether the set is closed, but in the assigned exercises, the set B is a closed set of measure zero. Because like you said, a rectangle Q is closed and bounded. – Seth Jan 12 '14 at 16:56
  • No, I am NOT saying, that any subset of a rectangle is compact, because it is false.

    But in the proof only compactness of $Q$ is used. This is actually the meaning of the remark ``The open rectangles ... may not cover $D$'' in Munkres

     – user68061 Jan 12 '14 at 17:54