Typewriter sequence of a U(0,1) random variable

This problem is a standard counterexample in measure theory in disguise:

Let {Z\sim\mathrm U(0,1)} and define

\displaystyle X_n = \mathsf 1_{[m2^{-k},\ (m+1)2^{-k})}(Z)

for {n=2^k+m}, {k\geqslant 1}, and {0\leqslant m<2^k}. In what sense does {X_n} converge to zero?

If {\varepsilon\in(0,1)} then

\displaystyle \mathbb P(X_n>\varepsilon) = \mathbb P(X_n=1)=2^{-k}.

Now, as {n=2^k+m}, we have {k=\lg(n-m)} and so {2^{-k}=\frac1{n-m}\leqslant\frac1n}. Hence

\displaystyle \mathbb P(X_n=1)\leqslant\frac1n\stackrel{n\rightarrow\infty}\longrightarrow0,

so that {X_n} converges to zero in probability.

It follows immediately that {X_n\stackrel{L^1}\longrightarrow 0}, as

\displaystyle \mathbb E[X_n] = \mathbb P(X_n=1)\leqslant\frac1n\stackrel{n\rightarrow\infty}\longrightarrow0,

and similarly {X_n\stackrel{L^p}\longrightarrow 0}, for {1\leqslant p<\infty} as

\displaystyle \mathbb E[X_n^p] = \mathbb P(X_n=1)\leqslant\frac1n\stackrel{n\rightarrow\infty}\longrightarrow0,

However, {X_n} does not converge almost surely to zero, as for {\varepsilon\in(0,1)} we have

\displaystyle  \begin{aligned} \mathbb P\left(\liminf_{n\rightarrow\infty} X_n<\varepsilon\right) &= \mathbb P\left(\bigcup_{n=2}^\infty\bigcap_{l=n}^\infty \{X_l=0\} \right)\\ &= \lim_{n\rightarrow\infty}\mathbb P\left(\bigcap_{l=n}^\infty \{X_l=0\} \right)\\ &= \lim_{n\rightarrow\infty}\mathbb P(Z<m2^{-k})\\ &= \lim_{k\rightarrow\infty}\mathbb P(Z<(n-2^k)2^{-k})\\ &\leqslant \lim_{k\rightarrow\infty}\mathbb P(Z<(2^{k+1}-2^k)2^{-k})\\ &=\lim_{k\rightarrow\infty}\mathbb P(Z<2^{-k+1})\\ &=0. \end{aligned}

The “disguise” is that {X_n} is really just the “typewriter sequence” from e.g. Terry Tao’s blog, with some minor tweaks that do not affect convergence.

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