Levi Convergence Theorem for step functions

**mikepol** · Jun24-09, 10:54 PM

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 $LaTeX Code: \\{s_n\\}$ be a sequence of step functions such that:

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

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

Then $LaTeX Code: \\{s_n\\}$ converges almost everywhere on

to a limit function

.

Proof:
Since integrals of $LaTeX Code: \\{s_n\\}$ converge, they must be bounded. Suppose that
$LaTeX Code: $\\lim_{n \\to \\infty} \\int_I s_n < M$$ . Let

be a set of points x of I such that

diverges. Given any $LaTeX Code: $\\epsilon$$ , define:
$LaTeX Code: B_n = \\{x | s_n(x) > \\frac{2M}{\\epsilon}\\}$
Then we have:

$LaTeX Code: B_n \\subseteq B_{n+1}$

$LaTeX Code: D \\subseteq \\bigcup_{n} B_n$

Also, each

is a finite collection of intervals. Define

to be the sum of the lengths of these intervals.

Now for any n:

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

So that $LaTeX Code: |B_n| < \\epsilon/2$ . From this we get that $LaTeX Code: $\\lim_{n\\to\\infty} |B_n| \\le \\epsilon/2$$ .
Since $LaTeX Code: |\\bigcup_{n=1}^m B_n| = |B_m|$ ,

is covered by a countable collection of intervals the sum of whose lengths is less than any $LaTeX Code: \\epsilon$ , therefore

has measure zero.
From this it follows that

is bounded almost everywhere, and so converges almost everywhere on I to some function

.