I am confused on the notations of step functions.
If we have a sequence of step functions $\{s_n : [0,1] \rightarrow \mathbb R \}_{n = 1}^{\infty}$ , what would $s_n(0)$ look like, as in if I plug zero in some function what would the steps to reach the output look like?

For example, take the step function $s_n = \sqrt n\ 1_{[0.5,0.5+\frac{1}{n}]}$, then for example what does $s_n(\frac{1}{2})$ mean?

Also what does it mean by $\lim_{n\to \infty}\int s_n(x) dx = 1$ and specifically how does that look like?

I am trying to imagine these graphically and I am confused. Apologies if this doesn't make sense. Any help we would be appreciated.

  • 1
    $\begingroup$ Search step function. Many pages and images there might help. $\endgroup$ Mar 18 at 1:56
  • $\begingroup$ I have tried but none of them seem to go into stuff like this, they more so prove propositions etc. $\endgroup$
    – liam song
    Mar 18 at 2:09

1 Answer 1


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 $g$.


$$s_n \left( \frac 1 2 \right) = \sqrt n \cdot \mathbf{1}_{[0.5,0.5+\frac{1}{n}]}\left( \frac 1 2 \right)$$

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

$$\begin{align*} \lim_{n \to \infty} \int_0^1 s_n(x) \, \mathrm{d} x &= \lim_{n \to \infty} \int_{1/2}^{1/2+1/n} s_n(x) \, \mathrm{d} x\\ &= \lim_{n \to \infty} \sqrt n \int_{1/2}^{1/2+1/n} \, \mathrm{d} x\\ &= \lim_{n \to \infty} \sqrt n x \bigg|_{1/2}^{1/2+1/n} \\ &= \lim_{n \to \infty} \frac{\sqrt n}{n} \end{align*} $$

  • $\begingroup$ This is what I was looking for. Thank you! $\endgroup$
    – liam song
    Mar 18 at 11:39

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