Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've come across some statements in measure theory that I don't fully understand. Consider the unit interval $I=[0,1]$ equipped with the Borel-$\sigma$-algebra $\mathcal{B}([0,1])$ and the Lebesgue measure. With $f$ an integrable function on $I$, $f_{n}$ is defined as follows $$f_{n}(x)=2^{n}\int_{(k-1)2^{-n}}^{k2^{-n}}f(y)dy,\ \ \ \ (k-1)2^{-n}\leq x\leq k2^{-n}$$

$\mathcal{F}_{n}$ is then the $\sigma$-algebra generated by the intervals of the form $[(k-1)2^{-n},k2^{-n})$ for $1\leq k\leq 2^{n}$.

It is clear to me that $\mathcal{F}_{n}$ is a filtration. But why is $f_{n}$ a martingale? I just don't see it. Could anyone help me see this?

After this, the statement is made that the upwards Lévy theorem can be applied to see that $f_{n}\rightarrow f$ a.s. and in $L_{1}$. I don't see how it can be applied to this case??

I've been struggling with this for a while now so help is much appreciated.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

If we show that $f_n=E[f\mid \mathcal F_n]$, we are done. To see this, notice that $f_n$ is $\mathcal F_n$-measurable. Let $I_{n,k}:=[(k-1)2^{-n},k2^{-n})$. It's enough to see that $\int_{I_{n,k}}f_nd\lambda=\int_{I_{n,k}}f\lambda$, which is by definition the case.

For the last part, we can show that the algebra $\bigcup_{n\geqslant 1}\mathcal F_n$ generated the Borel $\sigma$-algebra on the unit interval.

share|improve this answer
    
Oh thank you, haven't even thought about approaching it from that angle! –  user70267 Apr 5 '13 at 21:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.