Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Rudin - RCA p.20

Below statement is the definition of Lebesgue integral.

Let $(X,\mathfrak{M},\mu)$ be a measure space and $E\in \mathfrak{M}$ Let $f:X\rightarrow [0,\infty]$ be mesurable. Then there exists a simple measurable function $s:X\rightarrow [0,\infty)$ of the form $\sum_{i=1}^n \alpha_i \chi_{A_i}$ where $A_i=\{x\in X|s(x)=\alpha_i\}$ and $\chi$ is a characteristic function and $0≦s≦f$ and $a_i$ are mutually disjoint. Then $\int_E f d\mu \triangleq \sup (\sum_{i=1}^n \alpha_i \mu(A_i\cap E))$, the supremum being taken over simple functions.

=======================

Now, let $(X,\mathfrak{M},\mu)$ be a measure space and $E\in \mathfrak{M}$ and $s:X\rightarrow [0,\infty)$ be a simple measurable function of the form $\sum_{i=1}^n \alpha_i \chi_{A_i}$ where $A_i=\{x\in X|s(x)=\alpha_i\}$ and $a_i$ are mutually disjoint.

Here, how do i prove that $\int_E s d\mu = \sum_{i=1}^n \alpha_i \mu(A_i\cap E)$?

In other words, if $t:X\rightarrow [0,\infty)$ is a measurable simple function of the form $\sum_{i=1}^m \beta_i \chi_{B_i}$ where $B_i=\{x\in X|t(x)=\beta_i\}$ and $0≦t≦s$, how do i prove $\sum_{i=1}^n \alpha_i \mu(A_i\cap E) ≧ \sum_{i=1}^m \beta_i \mu(B_i\cap E)$?

Plus, i have heard many times that "Riemann integral is integrating over $x$-axis while Lebesgue integral is integrating over $y$-axis" (when a given function is a real function from real). However, if measurable space $X$ in the above definition is $\mathbb{R}$, this just seems 'fitting simple functions under a given graph of a function'. Why is this integrating over $y$-axis?

share|improve this question
    
@Jonas That was a typo. I'll edit it. Thank you –  Katlus Jan 10 '13 at 4:31
add comment

3 Answers

up vote 0 down vote accepted

It sounds like you've solved the problem; if not, I'll help out. For the question concerning Riemann vs. Lebesgue integration, I'll see if I can give you some motivation. Let's just pretend we're working with a smooth nonnegative function $f$ over $\mathbb{R}.$

In Riemann integration, we start by partitioning the $x$-axis, and then we capture the area under the curve by measuring how much of the $y$-axis can fit under the curve above a given element of our partition.

Conversely, with Lebesgue integration, we start to approximate $f$ by approximating the range of $f$ with the $\alpha_i$ used in the definition. In some sense, we're partitioning the $y$-axis into chunks that describe the $y$-behavior of $f$. Then, once we've approximately partitioned the range of $f$, we get a similar notion of area by measuring the $A_i$, which describe the sets of all $x\in \mathbb{R}$ for which the corresponding $\alpha_i$ is a 'good' approximation of $f$. You can see how, in a rough sense, Riemann integration gets area by chopping up $x$ and measuring $y$, and Lebesgue integration gets area by chopping up $y$ and measuring $x$. Only, in the latter case, we have better tools for describing measure.

The standard example is something like $f = \chi_\mathbb{Q}\cap [0,1]$. Obviously, the Riemann upper and lower sums are 1 and 0 respectively, so $f$ is not Riemann integrable, and partitioning the $x$-axis seems unfruitful. On the other hand, if we're using Lebesgue measure, we let $\alpha_0 = 0$ and $\alpha_1 = 1$ and apply your result to get $\int f = 0$. So this $y$-chopping and $x$-measuring lets us handle a wider variety of functions (in general).

The Wikipedia page on Lebesgue integration has a section for motivation/intuition, in case that's helpful.

share|improve this answer
    
Thank you, it really does help. –  Katlus Jan 10 '13 at 20:12
add comment

Note that $\sum_{j=1}^m \beta_j \mu(E\cap B_j)=\sum_{j=1}^m \sum_{i=1}^n \beta_j \mu(E\cap B_j \cap A_i)$.

share|improve this answer
    
I posted this Not as an answer. I thought this would be a proof, but I realized I still don't know why above inequality holds –  Katlus Jan 10 '13 at 5:53
    
Now it's done! It was hard to me to notice above equality.. –  Katlus Jan 10 '13 at 6:21
add comment

The is a consequence of monotone convergence theorem, which allows you to integrate step by step and the results are the same. I suggest you to read chapter 1 carefully, because I remember in the end he proved this statement himself.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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