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Let $(X, \mu)$ be a measure space. Let $X = A_1 \cup\cdots\cup A_k (A_i \cap A_j = \emptyset$ for $i \neq j)$, where each $A_i$ is measurable. We say $\pi = \{A_1,\dots,A_k\}$ is a finite measurable partition of $X$. We denote by $\mathfrak{P}$ the set of such partitions of $X$. Let $f\colon X \rightarrow [0, \infty]$ be a measurable function on $X$. Let $\pi = \{A_1,\dots,A_k\} \in \mathfrak{P}$. Let $a_i = \inf\{f(x); x \in A_i\}$. Let $b_i = \sup\{f(x); x \in A_i\}$.

We denote $\sum_i a_i\mu(A_i)$ by $\xi(f,\pi)$.

We denote $\sum_i b_i\mu(A_i)$ by $\Xi(f,\pi)$.

We denote $\sup\{\xi(f,\pi); \pi \in \mathfrak{P}\}$ by $\xi(f)$.

We denote $\inf\{\Xi(f,\pi); \pi \in \mathfrak{P}\}$ by $\Xi(f)$.

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


(1) $\xi(f) = \int_X f d\mu$.

(2) If $0 \le f \le M$, where $M$ is a constant such that $0 < M < \infty$, Then $\Xi(f) = \int_X f d\mu$.

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This seems correct, in fact very close to the standard definition. What is the definition you use? –  Alex Becker Aug 29 '12 at 1:09
@AlexBecker The one using simple measurable functions as defined in Rudin's Real and complex analysis. –  Makoto Kato Aug 29 '12 at 1:19
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1 Answer 1

We have $$ \xi(f,\pi)=\sum a_i\mu(A_i)=\sum\int_{A_i} a_i\leq\sum\int_{A_i}f=\int f, $$ and then $$\tag{1} \xi(f)\leq\int f. $$ If $g=\sum c_i\,1_{A_i}$ is any simple function with $g\leq f$, then $c_i\leq\inf\{f(x):\ x\in A_i\}$. So $\int g\leq\xi(f,\pi)$ for $\pi=\{A_1,\ldots,A_n\}$. This implies that $$\tag{2} \int f=\sup\left\{\int g\,:\ g \mbox{ simple with }g\leq f\right\}\leq\sup\xi(f,\pi)=\xi(f). $$ The inequalities (1) and (2) together show that $\int f=\xi(f)$.

The other equality is not true unless you require $f$ to be uniformly bounded. Indeed, consider $X=[0,1]$, $\mu$ the Lebesgue measure, $f=1/\sqrt x$. Then $\Xi(f)=\infty$ and $\int f=2$.

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Thanks. I corrected the proposition. –  Makoto Kato Aug 30 '12 at 20:49
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