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Lets consider the following two functions:

$$f(x)=\begin{cases} x &,x\in[0,1]\setminus\Bbb Q \\ 0 &,x\in[0,1]\cap\Bbb Q\end{cases}$$ $$g(x)=\frac{(-1)^{[x]}}{[x]}$$

where $[x]$ is the integer part of $x$. I am trying to determine which of these integrals are Riemann or/and Lebesgue integrable; $f(x)$ over $[0,1]$ and $g(x)$ over $[1,\infty)$.


Lebesgue for $f(x)$: $\int_{[0,1]} f dx=\int_{[0,1]\setminus\Bbb Q}x\ dx + \underbrace{\int_{[0,1]\cap\Bbb Q}0\ dx}_{=0}=\int_{[0,1]\setminus\Bbb Q}x\ dx=\int_{[0,1]}x\ dx-\underbrace{\int_{\Bbb Q}x\ dx}_{=0}=\frac12$

Riemann for $f(x)$: I assume this function is not Riemann integrable, because every interval in a partition of $[0,1]$ will contain both rationals and irrationals, but I am not sure how to conclude.


Riemann for $g(x)$: The function takes the values $-1$ on $[1,2)$, $\frac12$ on $[2,3)$, $-\frac13$ on $[3,4)$, etc. So the integral is $-1+\frac12-\frac13+\frac14-\ldots=\sum_{k=1}^\infty(-1)^k\frac1k=-\log(2)$.

But I have no idea for the Lebesgue integral of $g(x)$.

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$g$ is not lebesgue integrable, because the lebesgue integral of a function $g$ is defined as $$ \int_\Omega g \,d\mu := \int_\Omega g_+ \,d\mu - \int_\Omega g_- \,d\mu $$ where $g^+$ is the positive part of $g$, i.e. $g_+(x) = |g(x)|$ if $g(x) \geq 0$ and $g_+(x)=0$ otherwise, and $g_-$ simiarly is the negative part of $g$. $g$ is only integrable if at most one of the integrals on the RHS diverge. In your case, I believe that both $g^+$ and $g^-$ are $+\infty$.

$f$ is not riemann integrable because it's discontinuous everywhere.

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  • $\begingroup$ Is it a fact that the for the Riemann integral to exist the function can only be discontinuous at (at most) countably many points, and otherwise (discontinuous a.e.) the integral is never defined? $\endgroup$ – Phil-ZXX May 27 '13 at 11:01
  • $\begingroup$ @Thomas: No, the correct criterion (due to Lebesgue) is that for the Riemann integral to exist the function can only be discontinuous on a set of measure zero. $\endgroup$ – GEdgar May 27 '13 at 13:18
  • $\begingroup$ @Thomas When the riemann integral is introduced, it is often shown that if (but not only if!) there are only countably many discontinuities, then the riemann integral exists. The stronger statement that even a set of measure zero is OK requires a definition of measure, and is thus usually shown after the lebesgue integral has been introduced. The theorem is easy to image, though - basically, if the discontinuities for a set of measure zero, the terms in the riemann sums affected by them have smaller and smaller total length as the approximation gets better... $\endgroup$ – fgp May 27 '13 at 13:47
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For $f$: show that for any partition there is a Riemann sum that is zero, and the one which is bigger than, say, $\frac14$. As a result, the Riemann sums do not converge to the same value. For the Lebesgue integrability of $g$ you have to check whether at least one of the integrals $\int g^+$ or $\int g^{-}$ is finite. Otherwise, Lebesgue integral is not defined. Here $g^+ = \max(g,0)$ and $g^{-} = \max(-g,0)$

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  • $\begingroup$ So $g$ is not Lebesgue integrable, because, e.g. $\int g^+$ is just the sum $\sum_{k=1}^\infty \frac{1}{2k+1}$, which diverges? $\endgroup$ – Phil-ZXX May 27 '13 at 10:56
  • $\begingroup$ @Thomas: exactly $\endgroup$ – Ilya May 27 '13 at 12:40

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