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Let $[a, b]$ be a finite interval of the real line. A partition $P$ of $[a, b]$ is a finite sequence of numbers of the form

$a = t_0 < t_1 <\cdots < t_{k-1} < t_k = b$

Let $(X, \mu)$ be a measure space. Suppose $\mu(X) < \infty$. Let $f$ be a measurable function on $X$. Suppose $0 \le f(x) \le M$ for every $x \in X$, where $0 < M < \infty$.

Let $P\colon 0 = t_0 < t_1 <\cdots < t_{k-1} < t_k = M$ be a partition of $[0, M]$.

Let $A_i = \{x \in X; t_{i-1} < f(x) \le t_i\} (i = 1,\dots,k)$.

We denote $\sum_{i= 1}^k \mu(A_i)t_{i-1}$ by $s(f, P)$.

We denote $\sum_{i= 1}^k \mu(A_i)t_i$ by $S(f, P)$.

Let $\Phi$ be the set of partitions of $[0, M]$.

Let $s = \sup\{s(f, P); P \in \Phi\}$.

Let $S = \inf\{S(f, P); P \in \Phi\}$.

Is the following proposition true? If yes, how would you prove this?

Proposition

$s = S = \int_X f d\mu$.

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1 Answer 1

Yes. Note that $s(f,P)$ and $S(f,P)$ are the integrals of simple functions. These simple functions converge pointwise to $f$ along sequences archiving the supremum and infimum respectively. Furthermore, they are dominated by $M$, so the result follows from the dominated convergence theorem.

Details:

We have $s(f,P)=\sum_{i= 1}^k=\mu(A_i)t_{i-1}=\int\sum_{i= 1}^k 1_{A_i}t_{i-1}$, the integral of a simple function. Moreover, every such simple function is bounded in $[0,M]$ and is therefore dominated by the constant function with value $M$, which is integrable since we re working with a finite measure space. By the monotone convergence theorem, it suffices to show that there is a sequence $(P_n)$ such that $\lim_{n\to\infty} s(f,P_n)=s$ and the corresponding simple functions converge pointwise to $f$ (the case with the other approximation works similarly).

Note that if $P$ refines $P'$, then $s(f,P)\geq s(f,P')$. So we can take any sequence $(P_n)$ such that $\lim_{n\to\infty} s(f,P_n)=s$ and by refining each partition make sure that it converges in the same manner but such that no interval in the underlying partition is longer than $1/n$. But this means that the corresponding simple function is never more than $1/n$ different from $f$, so we can make the underlying simple functions converge to $f$.

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I think you are right. However, I prefer a more detailed and formal proof. Then, I think, more people can understand it. By taking the $\frac{1}{2^n}$ partition, I think we can get such a proof. –  Makoto Kato Aug 30 '12 at 20:37
    
@MakotoKato Feel free to post a more formal version of the same proof, different readers will appreciate different approaches. –  Michael Greinecker Aug 30 '12 at 23:58

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