Let $f_k\colon \mathbb{R}\longrightarrow \mathbb{R},\ k\in\mathbb{N}$ be given by
$f_k(x)=\frac{k}{\lfloor \sqrt{k^2x}\rfloor }\chi_{[4/k^2,1]}(x)$.
Note the floor function in the denominator, it is not displayed very clearly in the TeX code!
I want to compute $\lim_{k\to\infty}\int_{\mathbb{R}} f_k\mathrm{d}\lambda$.
Since $\{x\in\mathbb{R};\chi_{[4/k^2,1]}(x)>a\}=\begin{cases}[4/k^2,1]\text{ for }0\le a<1\\ \emptyset\text{ for } a\ge 1\\ \mathbb{R}\text{ for } a<0 \end{cases}$
i know that $f_k$ is measurable for every $k$.
I can't use monotone convergence theorem, since $(f_k)_{k\in\mathbb{N}}$ is not monotonically increasing. But for the integral
$\lim_{k\to\infty}\int_{\mathbb{R}}\frac{k}{\lfloor \sqrt{k^2x}\rfloor }\chi_{[4/k^2,1]}(x)\mathrm{d}\lambda(x)$ I am not entirely sure how to approach it best.
For $k>2$, you can write explicitly your function as $$ f_k(x) = \frac{k}{j} \quad \text{if}\ \ \frac{j^2}{k^2} \leq x < \frac{(j+1)^2}{k^2}, \quad j\in\{2, \ldots, k-1\}. $$ Indeed $f_k$ is different from $0$ only in the interval $I_k :=[4/k^2,1]$. Since $2 \leq \sqrt{k^2 x} \leq k$ for $x\in I_k$, you have that $[0,1) = \bigcup_{j=2}^{k-1} I_{k,j}$, with $$ I_{k,j} := \left[\frac{j^2}{k^2} , \frac{(j+1)^2}{k^2}\right), \qquad j \in \{2, \ldots, k-1\} $$ and $$ \lfloor \sqrt{k^2x}\rfloor = j \quad \forall x \in I_{k, j}, \quad \text{i.e.}\quad f_k(x) = \frac{k}{j} \quad \forall x \in I_{k, j}. $$ Hence $$ \int f_k = \sum_{j=2}^{k-1} \int_{I_{k,j}} f_k = \sum_{j=2}^{k-1} \frac{k}{j} \cdot \frac{(j+1)^2 - j^2}{k^2} = 2 \frac{k-2}{k} + \frac{1}{k} \sum_{j=2}^{k-1} \frac{1}{j} \to 2. $$