Calculating Lebesgue integral 2

Question

How can we calculate Lebesgue integral; for example, when we have $\mu(x)=x^2$ and $ f(x)=x $ $(\int f(x) \mu(dx) = \int x (dx)^2)$ ?

can we use $ \int f(x) \mu(dx)=\int fd\mu $ ,instead of solving $\int x (dx)^2$, calculate $\int x\times2x(dx)$ ?

Answer 1

So, the fundamental problem with your question is that you try to define a kind of integral of a function ($f$) with respect to something ($\mu$) which is not a measure. In a comment above I showed that $"\mu(x)=x^2"$ as a set function defined on intervals by

$$\mu([a,b))=(b-a)^2$$

was not a measure. The concept of the integral of functions with respect to set functions is restricted to specific set functions called measures.

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What is a measure and what does it have to do with probability? It is impossible to present measure theory here. It is easy to demonstrate, however, that if we want to model probabilities (of numbers falling into sets of reals) then we need special set functions.

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So, let $r_n(A)$ denote the relative frequency of an event that a random number fell in the set $A$. The most important property that we claim $r_n(A)$ to have are as follows:

1. $$1\ge r_n(A)\ge 0.$$
2. If $A$ and $B$ are disjoint sets then $$r_n(A\cup B)=r_n(A)+r_n(B).$$
3. $$r_n(\mathbb R)=1, \ \mathbb R \text{ being the real line}.$$

If the set function $\mu$ does not have these properties then it cannot be used as a model of the distribution of a real valued random variable.

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Your integral

$$\int_{\mathbb R} x\mu(dx)$$

is supposed to model the average of independent random trials. This is easy to (intuitively) see.

Imagine that $k_1, k_2,...k_m$ are the actual numbers that the random number fell into the pairwise disjoint sets $A_1, A_2,...A_m$ ($\bigcup_{k=1}^m=\mathbb R$), respectively. Also, let's select representative reals $x_1,x_2,...x_n$ from the sets $A_1, A_2,...A_m$ and let's agree that we will represent the random result by $x_k$ if the actual value of the random variable fell into the set $A_k$. If we had $n$ experiments then the average can be written as a sum:

$$\sum_{k=1}^m r_n(A_k)x_k\approx \int_{\mathbb R} x\mu(dx)$$

if $\mu$ as a set function well represents the probabilities of falling into corresponding sets.

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An example: Suppose that our random variable fall within the interval $[-1,1]$ surely. Then because of condition 3. we have to use a function generating a measure like the following one. Let

$$\mu(x)=\begin{cases} 0,& \text{ if } x<-1\\ \frac12 x^3+\frac12,& \text{ if } -1\le x\le1\\ 1,& \text{ if } x>1.\\ \end{cases}$$

Then, for an interval $[a,b]\subset [-1,1]$, let the probability of falling into $[a,b]$ be modeled by

$$\mu([a,b])=\frac12(b^3-a^3).$$

The expected value of our random phenomenon (an approximation of an actual average) is

$$\int_{-1}^1 x \ \mu(dx)=\int_{-1}^1 x \ \frac{\mu(dx)}{dx}\ dx=$$ $$\int_{-1}^1 x \ \frac{d (\frac12x^3+\frac12)}{dx}\ dx=\int_{-1}^1 x \ \frac32x^2\ dx=\frac32\int_{-1}^1 \ x^3 \ dx=0.$$

This argumentation was more than sketchy.

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If you want to understand these kinds of things better then start with : https://en.wikipedia.org/wiki/Measure_(mathematics).