# Prove the typewriter sequence does not converge pointwise.

1. Mar 24, 2013

### wrolsr

(Typewriter sequence) Consider the following sequence of functions on [0,1]. Let f1= X[0, 1\2], f2= X[1\2, 1], f3= X[0, 1\4], f4= X[1/4, 1\2], f5= X[1\2, 3/4], f6= X[3/4, 1], f7= X[0, 1\8], etc. Where X is the Characteristic function.

(a) Prove that fn does not converge for any x in [0,1].

(b) Show that this sequence of functions is a counterexample to the statement:
If for all n, fn and f are non-negative functions on E (with λ(E)<∞) and for all A contained in E, ∫A[\sub]fn dλ, then fn converges to f Lebesgue almost everywhere.

Attempt at solution: assume that lim n→∞fn(y)=0, which means that for all ϵ>0 there is an N, such that for all n > N, fn(y)<ϵ.

Last edited: Mar 24, 2013
2. Mar 24, 2013

### Dick

That's not much of an attempt. Can you explain in words why fn does not converge for any x in [0,1]?

3. Mar 24, 2013

### Robert1986

I suggest drawing the graph for each of the functions and you should be able to see what is happening and why fn(x) does not converge for any x.