# Levi Convergence Theorem for step functions

1. Jun 24, 2009

### mikepol

Hi Everyone,

I am currently reading Apostol's "Mathematical Analysis" to introduce myself to some advanced calculus techniques I will need in probability. Usually the proofs in this book are fairly straight forward, without too many missed steps, which is why I think this book is great for self study. I was reading about the Levi's convergence theorem for step functions and didn't like the proof that Apostol gives. To me it seems unnecessarily long and counter-intuitive. I think I have an alternative proof which is simpler to understand and shorter. But I started doubting myself, why would he present a long winded proof, when there is a simpler available - maybe something in my proof is wrong and I just can't spot it? That is why I want to share it here, maybe someone would be able to quickly see an error in my reasoning. Thank you very much in advance for anyone caring to taking a look.

Theorem:
Let $$\{s_n\}$$ be a sequence of step functions such that:

1). for every $$x \in I$$, $$s_{n+1}(x) \ge s_n(x)$$

2). $$\lim_{n \to \infty} \int_I s_n$$ exists.

Then $$\{s_n\}$$ converges almost everywhere on $$I$$ to a limit function $$f$$.

Proof:
Since integrals of $$\{s_n\}$$ converge, they must be bounded. Suppose that
$$\lim_{n \to \infty} \int_I s_n < M$$. Let $$D$$ be a set of points x of I such that $$s_n(x)$$ diverges. Given any $$\epsilon$$, define:
$$B_n = \{x | s_n(x) > \frac{2M}{\epsilon}\}$$
Then we have:

$$B_n \subseteq B_{n+1}$$

$$D \subseteq \bigcup_{n} B_n$$

Also, each $$B_n$$ is a finite collection of intervals. Define $$|B_n|$$ to be the sum of the lengths of these intervals.

Now for any n:

$$M > \int_I s_n \ge \int_{B_n} s_n > \int_{B_n}\frac{2M}{\epsilon} = \frac{2M}{\epsilon}\int_{B_n} 1 = \frac{2M}{\epsilon}|B_n|$$

So that $$|B_n| < \epsilon/2$$. From this we get that $$\lim_{n\to\infty} |B_n| \le \epsilon/2$$.
Since $$|\bigcup_{n=1}^m B_n| = |B_m|$$, $$D$$ is covered by a countable collection of intervals the sum of whose lengths is less than any $$\epsilon$$, therefore $$D$$ has measure zero.
From this it follows that $$s_n$$ is bounded almost everywhere, and so converges almost everywhere on I to some function $$f$$.

Last edited: Jun 25, 2009