2

Let $f:[0,1] \rightarrow \mathbb{R}$ be Riemann integrable.Prove that, if $\int_{[0,1]}f \neq 0$ then there exists $\delta>0$ & a non-empty open interval $I \subseteq[0,1]$ s.t $|f|\geq \delta$ throughout $I$.

Any hint on how to proceed? How do I construct such an open interval?

SL_MathGuy
  • 1,574
  • 1
    Try contradiction – Tab1e Jan 02 '20 at 00:13
  • If $f$ has a point $x_0$ where it is continuous and non-zero, then just by continuity you get such a $\delta=|f(x_0)|/2>0$ and an interval around it in which $|f|\geq \delta$. So, assume that at every point $x$ where the function is continuous one has $f(x)=0$. Since $f$ is Riemann integrable, then the set of points where $f$ is discontinuous has measure zero. Redefining $f$ at those points doesn't change the integral. So, $f$ and the function identically zero have the same integral. – MoonLightSyzygy Jan 02 '20 at 00:23
  • @MoonLightSyzygy "The set of points where $f$ is discontinuous has measure zero" is a result beyond the scope of this exercise, I believe. – Math1000 Jan 02 '20 at 00:34
  • 1
    @Math1000 I don't see the scope written anywhere. So, I solve it in the laziest way that I can find. – MoonLightSyzygy Jan 02 '20 at 00:36
  • Well the tags are "real-analysis" and "riemann-integration," not "measure-theory," "lebesgue-measure," or "lebesgue-integration." So that gives an idea of the scope. – Math1000 Jan 02 '20 at 00:38
  • @Math1000 Mathematics is not divided in pieces, but if anything Lebesgue criterion for Riemann integrability is real-analysis and covered in my undergraduate course for 1st year students. – MoonLightSyzygy Jan 02 '20 at 00:49
  • May be I should've added the tag 'measure theory'. @MoonLightSyzygy ,so the idea is to use the fact that, $f$ is R.I on $[a,b]$ iff $f$ is bounded & the set of points at which $f$ is discontinuous has measure 0? – SL_MathGuy Jan 02 '20 at 00:49
  • 1
    @SL_MathGuy This can probably be proven in lots and lots of ways. That is just the first way that popped in my head. – MoonLightSyzygy Jan 02 '20 at 00:51
  • We definitely didn't discuss Lebesgue measure while covering Riemann integration in my undergrad analysis course. My point is that there should be a more elementary way to proceed here. – Math1000 Jan 02 '20 at 01:02
  • 2
    This is equivalent to an exercise in Apostol's Mathematical Analysis. See https://math.stackexchange.com/a/2350319/72031 – Paramanand Singh Jan 02 '20 at 09:02

2 Answers2

2

Here is an elementary proof:

Take any partition $P=\{x_k\}^n_{k=1}$. Then, the Riemann sum $\sum^n_{k=1}f(x_k^*)\Delta x_k$ where $x_k^*$ is an $\textit{arbitrary}$ point in $[x_k,x_{k-1}]$ can always be chosen to equal zero because in every interval there is a point $x$ such that $f(x)=0$.

It follows that $\int^1_0f(x)dx=0$ if it exists.

Matematleta
  • 29,139
  • If you had planned to provide a mere hint rather than an answer, would you have considered the following? Let $f:[0,1] \rightarrow \mathbb{R}$ be Riemann integrable. Now prove the contrapositive of the corresponding part of the original problem. In other words, prove that: if there doesn't exist $\delta>0$ & a non-empty open interval $I \subseteq[0,1]$ s.t $|f|\geq \delta$ throughout $I$, then $\int_{[0,1]}f = 0$ – Ren Eh Daycart Jan 02 '20 at 13:34
  • 1
    @RenEhDaycart Yes, that is correct. – Matematleta Jan 02 '20 at 14:36
1

I will propose an elementary proof that depends only on Riemann sums (although in comments there is more simple and elegant one).

First step is to consider $|f|$. Since $f$ is Riemann integrable then $|f|$ is also Riemann integrable and moreover $0 < |\int\limits_0^1 f(x)dx| \le \int\limits_0^1 |f(x)|dx$. So, $\int\limits_0^1 |f(x)|dx > 0$.

In the second step we prove proposition by contradiction. Assume that for all open intervals $I \subset [0,1]$ and for all $\delta > 0$ there exists $x \in I$ s.t. $|f(x)| < \delta$. Now we consider lower Riemann (Darboux) sum that correspond to a partition $0 = x_0 < x_1 < \dots < x_n = 1$: $$s = \sum\limits_{i = 1}^n (x_i - x_{i-1})\inf\limits_{x \in [x_{i-1}, x_i]}|f(x)|. $$ From our assumption we have that $\inf\limits_{x \in [x_{i-1}, x_i]}|f(x)| = 0$. Therefore $s = 0$. But lower Riemann sums converge to $\int\limits_0^1 |f(x)|dx$ as partition diameter tends to $0$. Therefore $\int\limits_0^1 |f(x)|dx = 0$. That's a contradiction.

Matsmir
  • 2,545