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Let $A$ be a rectangle in $\mathbb{R}^2$. Suppose $f: A \to \mathbb{R}$ is bounded. Suppose $f(x) = 0 $ except on $F$, where $F$ is closed and has measure zero.

Does it follow that $f$ is integrable on $A$ and $\int_A f = 0 $ ?

Update with my try:

Suppose $|f(x)| \leq M$. Since $F$ is a closed(hence) compact and is of measure zero, we can cover $F$ with finite boxes, say $R_1,...,R_n$ such that $\sum_{i=1}^n Vol(R_i) < \frac{ \epsilon}{2M} $. We will construct a partition $P$ of $A$ that satisfy $U(f,P) - L(f,P) < \epsilon $ so that we will have $f$ is integrable on $A$. We can denote $R_i = [a_i,b_i] \times [c_i,d_i] $ for all $i=1,...,n$. We use all the $a_i,b_i,c_i,d_i$ to define a partition $P$ of $A$. Let $Q$ be an arbitrary subrectangle determined by the partition $P$. Since $f(x) = 0$ on every $Q \subset A \setminus F $, we only need to worry on the $Q's$ that satisfy $N \subset Q$. Hence

$$ U(f,P) - L(f,P) = \sum_{N \subset Q} ( \sup_{x \in Q} f - \inf_{x \in Q} f)Vol(Q) = \sum_{i=1}^n ( \sup_{x \in R_i} f - \inf_{x \in R_i} f) Vol(R_i) < 2M \frac{ \epsilon}{2M} = \epsilon $$

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Yes. Can you prove it? (Unfortunately, an answer must be at least 30 characters, hence this pointless text in parenthesis.)

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  • $\begingroup$ I have no idea how to prove it. Can you help me? $\endgroup$ – user145801 Apr 27 '14 at 9:39
  • $\begingroup$ Dear @MathcanbeFun, do you know that a bounded real-valued function on a rectangle is Riemann integrable if and only if its set of points of discontinuity has measure $0$? Also, can you prove that the integral of a bounded function on a set of measure zero is $0$? I would be very happy to discuss both of these points with you further if you are still stuck after thinking about them for a while. Note that the first point is a non-trivial theorem but the second part should be possible to prove as an exercise. $\endgroup$ – Amitesh Datta Apr 27 '14 at 9:43
  • $\begingroup$ I am trying to solve the problem right now using this fact that you mention. I will shortly update my question with my attempt, so you can give me some feedback if you don't mine. Thanks for your time! :) $\endgroup$ – user145801 Apr 27 '14 at 9:48
  • $\begingroup$ Ok I have made an attempt, though I still don't know how to show $\int_A f = 0 $. Is my attempt correct? $\endgroup$ – user145801 Apr 27 '14 at 10:08
  • $\begingroup$ Dear @MathcanbeFun, I think your attempt is correct. Do you know additivity of the integral: $\int_{X\cup Y} f = \int_{X} f + \int_{Y} f$ for disjoint measurable subsets $X$ and $Y$ of $\mathbb{R}^n$? $\endgroup$ – Amitesh Datta Apr 27 '14 at 10:29

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