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Define norm as $\int |f|$ (Riemann integral) on $\mathcal R^1[0,1]$, the space of riemann integrable functions on $[0,1]$ with identification $f=g$ iff $\int |f-g|=0$.

Let $\{ r_1,r_2,\cdots \}$ be the rationals in $[0,1]$, and let $f_n=1_{\{r_1,\cdots,r_n\}}$. Then $f_n$ is a cauchy sequence in $\mathcal R^1[0,1]$. I want to show that there is no $f\in \mathcal R^1[0,1]$ such that $f_n$ converges to $f$ in norm. How can I show it?

Obviously the pointwise limit $f=1_\mathbb{Q}$ is not contained in $\mathcal R^1[0,1]$, but can I use this fact? I think that there can be other candidates, since convergence in norm and pointwise convergence are different.

Gobi
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    I don't think your example works, you have $\int|f_n| = 0$ for every $n$ hence $f_n = 0$ and converges trivially. – Tim May 20 '13 at 16:27
  • @Tim Then I should find other examples. But then the above example is just saying that there is a (cauchy) sequence in $\mathcal{R}^1$ such that the pointwise limit is not in $\mathcal{R}^1$? – Gobi May 20 '13 at 16:31
  • Yes, but you can't say the $f_n$ are in $\mathcal R^1$ if you want a pointwise limit. Elements of $\mathcal R^1$ can't be evaluated at a point, so it's meaningless to talk about a pointwise limit. – Tim May 20 '13 at 17:10

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Recall the condition that $f=g$ if and only if $\int|f-g| = 0$. This means that elements of $\mathcal R^1$ are not functions in the classical sense, because they're only defined up to sets of measure $0$. You can't evaluate $f(x)$, because every pair $(x,y)$ there's some $g\in\mathcal R^1$ with $g=f$ but $g(x)=y$. We just change the value of $f$ at a single point.

So in your example we have $f_n = 0$ for every $n$.

Consider instead the functions

$$g_n(x) = \min\left(n,-\log x\right)$$.

Now each $g$ is Riemann integrable, and it's easy to see that the sequence is Cauchy $\int|g_n - g_m|\leq \int|g_n| - 1$ for $m>n$.

But there is no limit in $\mathcal R^1$. If there is a limit it must be $x\mapsto-\log(x)$ (almost everywhere), but that isn't Riemann itegrable because it's unbounded.

Dzoooks
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Tim
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    What about the lebesgue integral here: $"\int|\ln| d\lambda=\infty"$ – C-star-W-star Jul 14 '14 at 19:22
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    It seems as your example is not cauchy since the estimate rather explodes than vanishes, or? Though the idea is quite nice and elusive (take an unbounded integrable function and cut it appropriately to get it riemann integrable right?) – C-star-W-star Jul 14 '14 at 19:30
  • @Tim $-\log x$ is plain unbounded on $[0,1]$, you don't need to say almost everywhere. In fact, it's bounded almost everywhere! Anyway, I'm not sure what the motivation was for using that phrase in the first place. – Dzoooks Aug 08 '19 at 14:21
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    @C-Star-W-Star The sequence is indeed Cauchy; explicitly $$\int_0^1 g_m-g_n = e^{-n}-e^{-m} < e^{-n} \to 0.$$ But this example doesn't explain why Lebesgue integration is needed. $-\log x$ is unbounded, and so not Riemann integrable, but why not extend the Riemann integral to functions like $-\log x$ which are improper-ly integrable in the obvious way. Can someone point me to an example in which Lebesgue integration is truly needed? – Dzoooks Aug 08 '19 at 14:25
  • See copper.hat 's answer here. – Dzoooks Aug 08 '19 at 14:33
  • @C-Star-W-Star the logarithm is improperly integrable on $(0;1]$ (an antiderivative is $x\log(x)-x$). – reded Feb 03 '24 at 18:06