Limit of integrals

Question

In this question, I'm assuming the definition of Riemann Integrability.
a) Produce an example of a sequence $f_n \rightarrow 0$ pointwise on $[0,1]$ where $\lim_{n\to\infty}\int_0^1f_n$ does not exist.
b) Produce an example of a sequence $g_n$ with $\int_0^1g_n\to0$ but $g_n(x)$ does not converge to zero for any $x\in[0,1]$. Let's insist that $g_n(x)\geq 0$ for all $x$ and $n$.

For part (a), I thought of the function sequence $$f_n(x) = \begin{cases}1/x & \text{if }0<x<1/n\\0 &\text{if }x=0 \text{ or }x\geq1/n\end{cases}$$ This seems to work. I'm not sure about part (b) though. I would appreciate it if someone could check part (a) and offer some help with part (b).

Answer 1

For part $a$, your answer works, but none of the integrals exist. Try to see if you can alter your example so that each $f_n$ is integrable, but the limit of the integrals doesn't exist.

For $b$, try to think about each $g_n$ being an indicator function of a set; if the indicator functions get narrow quite slowly and slide across $[0,1]$, then the integrals will go to zero. It is possible to choose functions so that the $g_n$'s will not converge to zero pointwise.

Answer 2

I think part (a) is fine. But for part (b) how about this $$0\leq g_n(x)=\begin{cases}1+(-1)^n & x\in\{0,1\}\\1 &\frac k{2n}\leq x\leq\frac k{2n}+2^{-(n+1)}, k\in \Bbb{N}\\ 0 &(\mathrm{otherwise})\end{cases}$$

For each $n$, $g_n(x)$ equals $1$ over the intervals of the form $$\left[\frac k{2n},\frac k{2n}+2^{-(n+1)}\right]$$ where $k\in\Bbb{N}$ and $k<2n$, i.e., the intervals starting from multiples of $\frac 1{2n}$ with the length of $2^{-(n+1)}$. Note that there are $2n-1$ of such intervals within $[0,1]$. Notice that by this definition $g_n(x)$ does not have limit for any $x\in(0,1)$, because for a given $x$, as $n$ grows (changes) $x$ repeatedly falls in and out of the aforementioned intervals. Also, $g_n(0)=g_n(1)$ oscillates between $0$ and $2$, and thus $g_n(x)$ has no limit for any $x\in[0,1]$. Finally, for the integration limit we have $$\int_0^1g_n(x)\mathrm{d}x=(2n-1)\frac{1}{2^{n+1}}$$ which approaches zero as $n\to\infty$.