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Let $f$ be Riemann integrable and bounded on $[a,b]$. If $f(c) = 0$, at every point $c$ of continuity of $f$, then $\int^{b}_{a} [f(x)]^2 dx=0$.

Proof: This means, given $f$ is continuous at $c$ and $f(c)=0$, $\exists$ $(c-\delta, c+\delta) \subseteq [a,b]$ such that $ \left| f(x)-0 \right|<\epsilon, \forall x\in (c-\delta, c+\delta)$. Since $f\in R[a.b],$ is it true that $f$ (hence $f^2$) has infinitely many continuities? How do I use this to show that $L(f^2,P), U(f^2, P)< \epsilon$ ?

Appreciate any advice. Thank you.

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That's false. You're saying that if $f$ is continuous at $x$, and $f(x)=0$ then there is a small interval around $x$ on which $f\equiv 0$. Take $f(x)=x^2$ at $0$... –  xavierm02 Oct 5 '13 at 19:24
    
Thanks, I have made the necessary amendments. Hence, how should I proceed with the proof? –  Alexy Vincenzo Oct 5 '13 at 19:31
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Do you have the access to the result that the set of discontinuities of a Riemann integrable function has measure zero? –  Zach L. Oct 5 '13 at 19:39
    
And if you mean Riemann integrable, you should say that. –  GEdgar Oct 5 '13 at 19:42
    
@ZachL. No, I have not. –  Alexy Vincenzo Oct 5 '13 at 19:44

1 Answer 1

  1. Given $f$ is continuous at $c$ and $f(c)=0$, $\exists\, \delta=\delta(c,\epsilon)>0: (c−\delta,c+\delta) \subset [a,b]$ such that $|f(x)| < \sqrt{\epsilon} \, \forall x\in (c−\delta,c+\delta)$. (I have used the square root so that we can talk about the bounds on $f^2(x)$)
  2. Fix $\epsilon$, and represent $[a,b] = \cup_{c\in[a,b]: f(c) \in C^1} (c-\delta(c,\epsilon), c+\delta(c,\epsilon))$. If there are any intervals of $[a,b]$ that this cover misses, i.e., there are no continuity points in such an interval, then $f(\cdot)$ is not Riemann-integrable on these intervals. (You can show this using the standard argument for the indicator function of the rational numbers, together with the fact that $f$ is bounded: you can construct sequences of Darboux sums that would take the highest and the lowest possible values.)
  3. By Heine-Borel theorem, there exists a finite cover $[a,b] = \cup_{i=1}^{N_\epsilon} (c_i-\delta(c_i,\epsilon), c_i+\delta(c_i,\epsilon))$. Form the Darboux sum based on this finite number of intervals. The mesh of this partition is $\min \{\delta(c_i,\epsilon): i = 1, \ldots, N_\epsilon \}$, and that's where you crucially need the finiteness to have this minimum separated from zero. Now, for every $x\in (c_i-\delta(c_i,\epsilon), c_i+\delta(c_i,\epsilon))$, $f(x)^2 < \epsilon$, and thus this upper Darboux sum is less than $(b-a)\epsilon$. You can now go ahead and rescale everything from the outset if you like.

I admit that step 2 and argument about the number of discontinuity points is the weakest part of the whole thing.

I also think that you can have countably many discontinuities before you lose Riemann integrability. I believe that $$ f(x) = \left\{ \begin{array}{ll} 1, & x=2^{-n}, n\mbox{ is natural} \\ 0, & \mbox{otherwise} \end{array} \right. $$ is still Riemann integrable on say $[0,1]$. Say you have the grid defined on multiples of $2^{-k}$, i.e., $x_0=0,x_1=2^{-k}, 2\cdot2^{-k}, 3\cdot2^{-k}, \ldots, 1-2^{-k}, x_K=1$ for $K=2^k$. Then you can have at most $k+1$ intervals on which you can find $x$ such that $f(x)=1$, so your upper Darboux sum is $k2^{-k}\to0$ as $k \to \infty$.

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Thanks for the great explanation! I don't understand why $[a,b] = \cup_{c\in[a,b]: f(c) \in C^1} (c-\delta(c,\epsilon), c+\delta(c,\epsilon))$. Could you elaborate on that? What if $f$ has no points of continuity in $[a,c]$, where $c<b$ ? –  Alexy Vincenzo Oct 5 '13 at 20:16

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