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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.

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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?

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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.

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

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  • $\begingroup$ 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 $\endgroup$
    – Katlus
    Commented Jan 10, 2013 at 5:53
  • $\begingroup$ Now it's done! It was hard to me to notice above equality.. $\endgroup$
    – Katlus
    Commented Jan 10, 2013 at 6:21
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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.

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