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 \{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.
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.