Counting Measure Homework: Does fn(x) Converge?

[onthetopo]

Homework Statement

In the measure space {X,S,u} where u is the counting measure
X=(1,2,3,..}
S= all subsets of X
f_n(x)=$$\chi$$_{1,2,,,..n}(x) where $$\chi$$ is the characteristic (indicator) function.

Does f_n(x) converge
a.pointwise
b.almost uniformly
c.in measure

Homework Equations

The Attempt at a Solution

My guess would be
a.pointwise.yes, since it goes eventually to 1 , but it's hard to demonstrate this
b. almost uniformly : yes?
c. in measure: yes? follows directly from b if the answer to b is yes

 

[rochfor1]

a) Yes. Take $$x \in X$$. It shouldn't too difficult to show that eventually $$|f_n ( x ) - 1 | = 0$$.
b) See (c)
c) No. It's easy to see that the only possible limit is $$f\equiv1$$. Now, for any $$1>\varepsilon>0, n \in \mathbb{N}$$, $$\{ | f_n - 1 | > \varepsilon \} = \{ n + 1, n + 2, \ldots \}$$. What is the measure of this set? What does this say about almost uniform convergence?

 

[onthetopo]

a) Yes. Take $$x \in X$$. It shouldn't too difficult to show that eventually $$|f_n ( x ) - 1 | = 0$$.
b) See (c)
c) No. It's easy to see that the only possible limit is $$f\equiv1$$. Now, for any $$1>\varepsilon>0, n \in \mathbb{N}$$, $$\{ | f_n - 1 | > \varepsilon \} = \{ n + 1, n + 2, \ldots \}$$. What is the measure of this set? What does this say about almost uniform convergence?

I think the only problem is that we have to find the cardinality of (n+1,n+2,n+3...) as n goes to infinity. As n goes to infinity, there is no number larger than n and in fact no n+1,n+2...can exist?

 

[rochfor1]

That's not a very precise way to think about it. Think about it...without a doubt $$n\to\infty$$, but at any "stage" of this limit, $$n<\infty$$ so the set I wrote about is well-defined, and can in fact be mapped bijectively to the natural numbers. What does that imply about its cardnality?

 

[onthetopo]

I completely understand the solution now.

One last question
How to prove that the metric space (L,d)
where L=all measurable functions
d(f,g)=$$\int{\frac{|f-g|}{1+|f-g|}$$ is complete?

I really have no idea how to show this since I have to show that the limit of ARBITRARY cauchy sequence is another measurable function. I think it is very difficult.

 

[Dick]

Is that the complete question? What domain are your measurable functions defined on?

 

[onthetopo]

Sorry, my bad (X,S,u) is a FINITE measure space and L is the set of FINITE measurable functions. no information other than that.
no mention of lebesgue measure or borel set

 

[Dick]

That helps. Otherwise you wouldn't have a metric. Now your metric is equivalent to the L_1 metric, isn't it? Sorry, it's been a long time since my Real Analysis classes and I don't have this stuff at the tip of my tongue anymore.