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.
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 $0<a<1$)