Given $f_{k}(x) = k\mathbf{1}_{(0,\frac{1}{k}]}(x)$ I wanted to compute the limit as $k \to \infty$.
Is the following correct:
Let $x \in \mathbb{R}$ such that $x \notin (0, \frac{1}{k}]$ then for $N \in \mathbb{N}$, sufficiently large (e.g. $N > x$) we have $\mathbf{1}_{(0, \frac{1}{k}]}(x) = 0$ and this holds for all $n \geq N$.
Therefore the pointwise limit is $0$.
Note: As per comments this has been addressed as not being correct and I have amended it.
Follow up question:
If I calculate the $\lim_{k \to \infty}\int_{\mathbb{R}}f_{k} d\mu $ and $\int_{\mathbb{R}}\lim_{k \to \infty}f_{k}d\mu$ where $d\mu$ is Lebesgue measure I obtain two different values. Graphing the function it is clear that the dominated convergence theorem cannot apply, because there is no "dominating" function. I'm assuming (since I get two different values), that the monotone convergence theorem also cannot apply, but I can't work out why. From what I can see, $f_{k}(x) \leq f_{k+1}(x), \forall x$ and we know that the pointwise limit exists…
Best Answer
The pointwise limit is indeed $0$ for every $x$ but your "proof" is opaque when it comes to the use of indices $k,N,n$.
The intention might be okay, but I would not classify it as a correct proof.
The essence is that it can be proved that for every $x\in\mathbb R$ some positive integer $n$ exists with: $$k\geq n\implies x\notin(0,\frac1k)$$
That implies directly that for every $x\in\mathbb R$ some positive integer $n$ exists with: $$k\geq n\implies f_k(x)=0$$ hence for every $x\in\mathbb R$: $$\lim_{k\to\infty}f_k(x)=0$$
edit:
We do not have $f_k\leq f_{k+1}$ here because for $x\in\left(\frac1{k+1},\frac1{k}\right]$ we have $f_{k+1}(x)=0<f_k(x)$.