Let $f:[a,b] \rightarrow\mathbb{R}$ be Riemann integrable and U=$\{x \in [a,b]|f(x)\neq0\}$ have zero measure. Prove that $\int_a^bf(x)=0.$
I'm a bit lost since $U$ isn't closed and thus isn't compact, so how do I build a partition for $[a,b]$?

Here's my attempt at it:Let $U = \bigcup_{n\geq1}E_n$, where $E_n=\{x\in[a,b]| |f(x)| \geq 1/n\}$. Since $E_n \subset U$, $m(E_n)=0, \forall n$. Note also that $E_n$ is closed and thus compact.
Thus given $\epsilon >0$, choose $n$ such that $\frac{b-a}{n} < \epsilon$. Since $m(E_n) = 0$ there exists open interval $I_1,I_2,...$ such that
$E_n \subset \bigcup I_j, \Sigma|I_j|<\frac{\epsilon}{2(M-m)}$, where $M=\sup f,m=\inf f$.
Since $E_n$ is compact there exists a finite subcover $I_1,...,I_n$. Take the partition $P$ such that it's points are extremes of $I_1\cap[a,b],...,I_n\cap[a,b]$. 4Therefore:
$$S(|f|,P)=\Sigma M_i(x_i-x_{i-1})+\Sigma M_i (x_i-x_{i-1})$$ Where the first sum is taken on intervals that do not intercept $E_n$ and the second on intervals that do. Thus
$$S(|f|,P)\leq\Sigma(1/n)(x_i-x_{i-1})+\Sigma(M-m)(x_i-x_{i-1})$$ $$S(|f|,P)\leq\frac{b-a}{n}+(M-m)\frac{\epsilon}{M-m}$$ $$S(|f|,P)<\epsilon/2+\epsilon/2=\epsilon$$ Thus $\overline\int|f|=0\implies\overline\int f=0\implies\int f =0$
Is this ok?

  • 2
    $\begingroup$ is this a Riemann or lebesgue integral? $\endgroup$ – qbert Mar 13 '18 at 22:18
  • $\begingroup$ The phrase "zero measure" strongly suggests that this integral is in the Lebesgue sense. $\endgroup$ – probably_someone Mar 13 '18 at 22:21
  • $\begingroup$ @probably_someone when I learned the Riemann integral, we discussed discontinuities on sets of measure zero and how they did not affect the value of the integral $\endgroup$ – qbert Mar 13 '18 at 22:21
  • $\begingroup$ Well then, there you go. Just choose your partition to be all of the segments that don't include the points in $U$. $\endgroup$ – probably_someone Mar 13 '18 at 22:22
  • 2
    $\begingroup$ @probably_someone On the other hand, "partition" sounds like Riemann $\endgroup$ – zhw. Mar 13 '18 at 22:30

Sketch: I assume $f$ is Riemann integrable on $[a,b]$ and you want to stay within this context. Let $P$ be a partition of $[a,b].$ Then $P$ induces subintervals $I_k, k =1,2,\dots ,n.$ Because $U$ has zero measure, $U$ does not contain any $I_k.$ So for each $k$ there exists $c_k\in I_k$ such that $f(c_k)=0.$ It follows that

$$\sum_{k=1}^{n} f(c_k)\Delta x_k=0.$$

Since $\int_a^b f$ is the limit of such sums (as the mesh size of $P \to 0),$ we have $\int_a^b f=0.$

  • $\begingroup$ Nothing could beat this in terms of simplicity. +1. There are occasions where Riemann sum is handy compared to a Darboux sum. $\endgroup$ – Paramanand Singh Mar 14 '18 at 14:30

Just to cover our bases, I will offer a solution for the Lebesgue integral.

Since $U\cup [a,b]\setminus U=[a,b]$ is a disjoint union, we know that the integral is additive on these sets and $$ \int_{[a,b]}f=\int_{U}f+\int_{[a,b]\setminus U}f=0 $$ since $U$ is measure zero and $f$ vanishes on $[a,b]\setminus U$.


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