Give an example of a sequence of functions on $[0,1]$ with the property that $\lim_{ n→∞} f_n(x) = 0$ for all 0 ≤ x ≤ 1 and yet for every interval $[c,d] ⊂ [0,1]$ and every N there is some x ∈ [c,d] and n > N with $f_n(x) > 1$.
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Look at 1. from Saz's answer to this question: Convergence types in probability theory : Counterexamples.
The idea of this, and any other such counterexample, is that you can move around the points $y$ where $f_n(y) > 1$ as $n$ increases such that each point can have limit $0$ (because for any given point $x$ $f_n(x)>1$ only finitely many times) but no matter how large $n$ is there is always at least some point $z$ for which $f_n(z) >1$ even if for most points the opposite is true. Thus this possibility depends in large part on the fact that $[0,1]$ is an infinite set.
Chill2Macht
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please clarify the definition of the random variables – Dhrubajyoti Sarkar Aug 13 '16 at 18:22
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1@jaggu Do you know what an indicator function is? $1_A(x) = \begin{cases} 1 & x \in A \ 0 & x \not\in A \end{cases}$ – Chill2Macht Aug 13 '16 at 18:31
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1the set of points where $X_n$ vanishes contains the points of the form $\frac{m}{2^n}$ for $m,n \in N$. Am I right? – Dhrubajyoti Sarkar Aug 13 '16 at 19:32
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@jaggu yes you are. – Chill2Macht Aug 13 '16 at 19:40
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Then what about the $\lim X_n$? – Dhrubajyoti Sarkar Aug 13 '16 at 19:41
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Take $\{q_n\}_{n\geq 0}$ be the set rational numbers in $[0,1]$ then let $$f_n(x)=\begin{cases} 2\quad\mbox{if $x=q_m$ for $m\geq n$},\\ 0\quad\mbox{otherwise.}\\ \end{cases}$$ Clearly for all $x\in [0,1]$, $\displaystyle \lim_{ n→∞} f_n(x) = 0$. Moreover, for every interval $[c,d]\subset [0,1]$, since the sequence $\{q_n\}_{n\geq N}$ is dense for any $N$, there is a $x:=q_n\in [c,d]$ for $n>N$ and by definition $f_n(x)=2>1$.
Robert Z
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1This might be worth noting: There is no sequence of continuous functions with this poperty. – zhw. Aug 13 '16 at 19:40