3
$\begingroup$

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?

Proposition

(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 $\Xi(f) = \int_X f d\mu$.

$\endgroup$
  • 1
    $\begingroup$ This seems correct, in fact very close to the standard definition. What is the definition you use? $\endgroup$ – Alex Becker Aug 29 '12 at 1:09
  • $\begingroup$ @AlexBecker The one using simple measurable functions as defined in Rudin's Real and complex analysis. $\endgroup$ – Makoto Kato Aug 29 '12 at 1:19
3
$\begingroup$

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 $\int f=2$.

$\endgroup$
  • $\begingroup$ Thanks. I corrected the proposition. $\endgroup$ – Makoto Kato Aug 30 '12 at 20:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.