This is an exercise of the course of Measure and Integration and I'm having trouble to solve this. I not know how to show the sequence is of Cauchy and why are not R-Integrable.

Let the sequence of functions $f_n:[0,1] \rightarrow \mathbb{R}$ such that:$$ f_n(x)= \left\{\begin{array}{lll}0& \textrm{, if }&0\leq x \leq 1/(n+1)\\n^{1/2}((n+1)x-1)& \textrm{, if }& 1/(n+1)\leq x \leq 1/n\\1/\sqrt{x}&\textrm{, if }& 1/n\leq x \leq 1\\\end{array}\right.$$ Show that $\{f_n\}$ is a Cauchy sequence in $C^0([0,1])$, relative to metric : $$d(f,g)=\int_0^1|f(x)-g(x)|dx.$$

Show that $f_n$ converge pointwise for the function $f:[0,1] \rightarrow \mathbb{R}$ and this function is not Riemann integrable.

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    $\begingroup$ $C^{0}([a,b])$ might denote continuous functions $[0,1]\to \mathbb{R}$. $\endgroup$ – T. Eskin Jan 31 '14 at 14:06
  • $\begingroup$ @ThomasE. That's the problem with having done a lot of exercises speaking of the Cantor set. Thanks, already excludes this of the original question. $\endgroup$ – Felipe Jan 31 '14 at 14:09
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    $\begingroup$ For integrability, note the pointwise limit satisfies $f(x)=1/\sqrt x$ for $x\ne0$. This is not bounded on $[0,1]$. $\endgroup$ – David Mitra Jan 31 '14 at 14:12

To show the sequence is Cauchy, you first of all should compute $d(f_n,f_m)$. If we assume wlog. that $m>n$, then we have according to the piecewise definition $$\begin{align}d(f_n,f_m)=\int_0^1|f_n(x)-f_m(x)|\,\mathrm dx= \int_0^{\frac1{m+1}}&|f_n(x)-f_m(x)|\,\mathrm dx \\+\int_{\frac1{m+1}}^{\frac1{m}}&|f_n(x)-f_m(x)|\,\mathrm dx\\+\int_{\frac1{m}}^{\frac1{n+1}}&|f_n(x)-f_m(x)|\,\mathrm dx\\+\int_{\frac1{n+1}}^{\frac1{n}}&|f_n(x)-f_m(x)|\,\mathrm dx\\+\int_{\frac1{n}}^{1}&|f_n(x)-f_m(x)|\,\mathrm dx\end{align} $$ This looks a bit complicated, but can be treated piecewise. For Cauchy you need to show that $d(f_n,f_m)$ with $m>n$ is bounded by an expression that depends on $n$ only and that tends to $0$ as $n\to \infty$.

The pointwise limit of the $f_n$ is (as should be obvious) $$ f(x) = \begin{cases}0&\text{if }x=0\\\frac1{\sqrt x}&\text{if }0<x\le 1\end{cases}.$$ Do yo see why that is not Riemann integrabel? (Hint: You can compute $\int_a^1\frac{\mathrm dx}{\sqrt x}$ for $0<a<1$)

  • $\begingroup$ I'm lost here: $$|f_m -f_n| = d(f_m,f_n)=\int_0^1 |f_m(x)-f_n(x)| dx=$$$$=\int_0^{1/(n+1)} |f_m(x)-f_n(x)| dx+\int_{1/(n+1)}^{1/n} |f_m(x)-f_n(x)| dx+\int_{1/n}^{1} |f_m(x)-f_n(x)| dx=$$$$=\int_0^{1/(n+1)} |0-0| dx+\int_{1/(n+1)}^{1/n} |m^{1/2}((m+1)x-1)-n^{1/2}((n+1)x-1)| dx+ $$$$+\int_{1/n}^{1} |1/\sqrt{x}-1/\sqrt{x}| dx=\int_{1/(n+1)}^{1/n} |m^{1/2}((m+1)x-1)-n^{1/2}((n+1)x-1)| dx=$$$$=\int_{1/(n+1)}^{1/n} |(m^{1/2}((m+1)-n^{1/2}(n+1))x+n^{1/2}-m^{1/2})| dx= $$$$=\left. |(m^{1/2}((m+1)-n^{1/2}(n+1))\frac{x^2}{2}+x(n^{1/2}-m^{1/2})|\right |^{1/n}_{1/(n+1)} $$ Can you help-me? $\endgroup$ – Felipe Jan 31 '14 at 15:58

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