To say that
$$s_n = \sqrt n \cdot \mathbf{1}_{[0.5,0.5+\frac{1}{n}]}$$
is to say that
$$\text{for each $x \in [0,1]$, we define } s_n(x) := \sqrt n \cdot \mathbf{1}_{[0.5,0.5+\frac{1}{n}]}(x)$$
Note (Characteristic/Indicator Functions): For a set $S \subseteq A$, the function $\mathbf{1}_S : A \to B$ (sometimes denoted $\chi_S$) is defined by the rule
$$\mathbf{1}_S(x) := \begin{cases} 1, & x \in S \\ 0, & x \not \in S \end{cases}$$
A less-complicated notion may be defining functions $f,g$ by, say,
$$f = \cos \qquad g = 2 \exp$$
which identically means
$$f(x) = \cos(x) \qquad g(x) = 2 \exp(x) = 2 e^x$$
for whatever domains we attribute to $f$ and $g$.
Hence,
$$s_n \left( \frac 1 2 \right) = \sqrt n \cdot \mathbf{1}_{[0.5,0.5+\frac{1}{n}]}\left( \frac 1 2 \right)$$
Of course, for every $n \in \mathbb{N}$, we have $\frac 12 \in \left[\frac 12,\frac12 + \frac 1n\right]$, so the characteristic function evaluates to $1$ and
$$s_n \left( \frac 1 2 \right) = \sqrt n $$
On the other hand, $0$ is never in any such interval, so
$$s_n(0) = 0$$
for each $n$.
A visual for what these $s_n$ look like is on Desmos here. Note that each $s_n$ is its own, unique function.
The limit is more self-explanatory. Presumably, the integral is meant to be over $[0,1]$, i.e. you seek
$$\lim_{n \to \infty} \int_0^1 s_n(x) \, \mathrm{d} x$$
There is nothing unusual here or different from what you saw in ordinary calculus classes. You have a quantity dependent on $n$ (the integral); find the limit of that quantity as $n \to \infty$.
As a note, since $s_n(x) = 0$ when $x < \frac 1 2$ or $x > \frac 1 2 + \frac 1 n$, we may write
$$\begin{align*}
\lim_{n \to \infty} \int_0^1 s_n(x) \, \mathrm{d} x
&= \lim_{n \to \infty} \int_{1/2}^{1/2+1/n} s_n(x) \, \mathrm{d} x\\
&= \lim_{n \to \infty} \sqrt n \int_{1/2}^{1/2+1/n} \, \mathrm{d} x\\
&= \lim_{n \to \infty} \sqrt n x \bigg|_{1/2}^{1/2+1/n} \\
&= \lim_{n \to \infty} \frac{\sqrt n}{n}
\end{align*}
$$
