If $f : [a,b] \to [0,M]$ is Riemann integrable and has integral zero then $f(x) = 0$ at every continuity point $x$ of $f$.

I don't know how to prove this though I have faint idea stated below. Can we use this idea?


We are given that function is Riemann integrable and integral is zero.

$$\lim_{mesh \space P \to \space 0} R(f,P,T) = I$$

So, Given $\epsilon > 0, \exists\delta>0, $ such that for any partition pair $$mesh \space P < \delta \implies \left| R(f,P,T) - I\right| < \epsilon $$

If number of intervals are n then $$ \left| \sum_{i=1}^n f(t_i)\Delta x_i \right| < \epsilon$$

Now we can take contradiction that function has value greater than 0 at some continuous points and then use this to contradict : If function is Riemann integrable, then it's set of discontinuities will be zero set.

But I am not able to formulate it. Give me hint. Thanks.

  • $\begingroup$ You can use a theorem that states that $f$ is Riemann integrable if and only if the set $D$ of points at which $f$ fails to be continuous has measure zero in $\mathbf{R}^n$? $\endgroup$ – positrón0802 Jul 22 '17 at 4:53
  • $\begingroup$ @positrón0802 But I don't know how to use it. $\endgroup$ – MeetR Jul 22 '17 at 4:54
  • $\begingroup$ @positrón0802 What you stated is a deep theorem due to Lebesgue and is not related to the current question. We are not asked to prove that $f=0$ a.e.. We are just asked to proved that $f(x_0)=0$ if f is continuous at $x_0$. $\endgroup$ – Danny Pak-Keung Chan Jul 22 '17 at 5:08
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    $\begingroup$ It is rather odd that your book presents this as a corollary of a deep and difficult theorem when the result is easily available via contradiction as shown in @DannyPak-KeungChan answer. $\endgroup$ – Paramanand Singh Jul 22 '17 at 5:13
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    $\begingroup$ However I don't think the question deserves downvotes. +1 to compensate. $\endgroup$ – Paramanand Singh Jul 22 '17 at 5:17

Prove by contradiction. Suppose that there exists $x_{0}\in[a,b]$ such that $f$ is continuous at $x_{0}$ and $f(x_{0})>0$. Let $c=\frac{1}{2}f(x_{0})>0$. Since $f$ is continuous at $x_{0}$ and $f(x_{0})>c$, there exists $\delta>0$ such that $f(x)>c$ for any $x\in[x_{0}-\delta,x_{0}+\delta]$. (We need to adjust the interval $[x_{0}-\delta,x_{0}+\delta]$ accordingly if $x_{0}=a$ or $x_{0}=b$.)

Since $f\geq0$, we have $$ \int_{a}^{b}f(x)\,dx\geq\int_{x_{0}-\delta}^{x_{0}+\delta}f(x)\,dx\geq\int_{x_{0}-\delta}^{x_{0}+\delta}c\,dx=2\delta c>0, $$ which is a contradiction.

  • $\begingroup$ For completeness' sake: With the mentioned adjustment for the boundaries, it is $$\int_a^bf(x)\,\mathrm dx\ge\int_{\max\{a,x_0-\delta\}}^{\min\{b,x_0+\delta\}}f(x)\,\mathrm dx\ge \int_{\max\{a,x_0-\delta\}}^{\min\{b,x_0+\delta\}}c,\mathrm dx\ge \min\{\delta,b-a\}c>0$$ $\endgroup$ – Hagen von Eitzen Jul 22 '17 at 6:08

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