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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.

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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.

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

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