How to calculate the limit of the characteristic functions $\chi_{[0,n]}$ and $\chi_{[-n,n]}$

Question

Hi I am unsure of how you would evaluate the limit as $n\longrightarrow \infty$ of characteristic functions such as $\chi_{[0,n]}$ $\chi_{[-n,n]}$. Would their limits simply be $\lim_{n\to\infty} \chi_{[0,n]}=\chi_{[0,\infty)}$ and $\lim_{n\to\infty} \chi_{[-n,n]}=\chi_{(-\infty,\infty)}$ or are they something else?

Answer 1

You have to be a bit careful when talking about the limits of sequences whose elements are functions and not points. We would say that $\chi_{[0,n]}$ converges to $\chi_{[0,\infty)}$ pointwisely since for every $x \in [0,\infty)$ and for $\varepsilon > 0$, certainly there is $n$ large enough so that $n > x$ and thus $|\chi_{[0,n]}(x) - \chi_{[0,\infty)}(x)| = 0 < \varepsilon$. However, this convergence is not uniform since for any $n$ there is $x > n$ so that $|\chi_{[0,n]}(x) - \chi_{[0,\infty)}(x)| = 1 > \varepsilon$ for sufficiently small $\varepsilon$. You can see how a similar line of reasoning might apply to $\chi_{[-n,n]}.$

Answer 2

The indicator function is compatible with $\varliminf$ and $\varlimsup$:

• $\varliminf 1_{A_n}=1_{\varliminf A_n}$
• $\varlimsup 1_{A_n}=1_{\varlimsup A_n}$

With these defined for sets as $\begin{cases} \varliminf A_n=\bigcup\limits_{n\in\mathbb N}\bigcap\limits_{k=n}^\infty A_k\\ \varlimsup A_n=\bigcap\limits_{n\in\mathbb N}\bigcup\limits_{k=n}^\infty A_k\end{cases}$

And we say the sequence of sets has a limit $\lim\limits_{n\to\infty} A_n$ when both limits are equal.

In that case the indicator function also has a pointwise limit which is given by the expected $1_{\lim\limits_{n\to\infty} A_n}$.

Note that the above condition is important, for instance $\big(\frac{(−1)^n}n, 1 − \frac{(−1)^n}n\big]$ has a lim inf which is $(0,1)$ but the lim sup $[0,1]$, so it is delicate to give a value for the indicator limit in points $0$ and $1$.

For your examples, this poses no issue since your $A_n=[0,n]$ and $B_n=[-n,n]$ have proper limits as sets, resp. $[0,\infty)$ and $\mathbb R$.

But as functions, they do not converge for $||\cdot||_\infty$ norm, as Kyle Duffy stated in his answer.