Convergence of Sequence of Indicator Functions

Question

With $I \subset \mathbb{R}$, let $\chi_{i}$ denote the indicator function of I. Define a sequence of functions $\{g_{n}\}_{n=1}^{\infty}$ on $[0,1]$ as:

$g_{1}(x) = \chi_{[0,1]}(x),\quad g_{2}(x) = \chi_{[0,1/2]}(x),\quad g_{3}(x) = \chi_{[1/2,1]}(x),\\$ $g_{4}(x) = \chi_{[0,1/4]}(x),\quad g_{5}(x) = \chi_{[1/4,1/2]}(x),\quad g_{6}(x) = \chi_{[1/2,3/4]}(x),\quad g_{7}(x) = \chi_{[3/4,1]}(x) $

and so on.

How can I show that $\{g_{n}\}_{n=1}^{\infty}$ converges at no point in [0,1]?

Answer 1

Hint

For $x \in [0,1]$ you'll find integers $n$ as large as you want such that $g_n(x)=0$ and for others, $g_n(x)=1$. Hence $\{g_n(x)\}$ cannot be a convergent sequence.

Answer 2

Consider, the subsequence $(g_1(x), g_2(x), g_4(x), .....g_{2^n}(x).....) \to 0$ but the subsequence $(g_1(x), g_3(x), g_7(x),....., g_{2^n-1}(x).....) \to 1$ .