show that $g_k = kI_{(0,1/k)}$ but that $\lim \,_{(k\rightarrow \infty)} \int f_k d\mu \neq 0$

Question

I do not know how to start prove this definition in one of text book. Someone please show it to me. any HINT really appreciated..

Let $S=[0,1]$, $\Sigma=\mathcal{B}([0,1])$ and $\mu$ Lebesgue measure. Define $g_k = kI_{(0,1/k)}$. How to prove that $g_k(s) \rightarrow 0$ for every $s \in S$, but that $\lim _{k\to\infty} \int g_k d\mu \neq 0$.

Many thanks for your help..

Answer 1

If $x \in (0, 1]$ is given, choose an $N$ for which $\frac{1}{N} < x$. Then whenever $n \ge N$, we see that $$\frac{1}{n} \le \frac{1}{N} < x \implies x \notin (0, \frac{1}{n}) \implies g_k(x) = 0$$

If $x = 0$, then $g_k(x) = 0$ for every $k$. This establishes the first claim by definition of sequence convergence.

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On the other hand, each $g_k$ is a simple function, so its Lebesgue integral is easily computed by

$$\int g_k dm = k m\Big((0, \frac{1}{k})\Big) = \frac{k}{k} = 1$$

as desired.