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0.1 “Limit”——footstone of mathematical analysis

\[if\ x_{n}\longrightarrow x\ \ \Rightarrow f(x_{n})\longrightarrow f(x)\]

So we can see the importance of continuous fuctions in Mathematical Analysis.

Assume there are two metric space \((E,\ d)\) , and \((S,\ \rho)\).

\(f:\ E\longrightarrow S\) then the following proposition are equivalent:

  1. The definition of continuous functtion that most of us learnt in Mathematical Analysis:

\[\forall\epsilon>0,\ \exists\delta>0,\ such\ that\ f(B_{d}(x_{0},\delta))\subset B_{\rho}(f(x_{0}),\epsilon)\]

Remark: Most of us are used to the definition in \((R,\ d)\). However, it is only a case of metric space. Though it helps to understand by realine \(R,\) or \(R^{2}\), \(R^{3}\)……, metric space is much broader than that.

  1. \(\forall G\subset S\), \(G\) is open, we have \(f^{-1}(G)=\{x\in E:\ f(x)\in G\}\) is open subset of \(E\)
  2. \(\forall F\subset S\), \(F\) is closed, we have \(f^{-1}(F)=\{x\in E:\ f(x)\in F\}\) is closed subset of \(E\)

0.2 Riemann integral and Lebesgue integral

We learnt in Mathematical Analysis about Riemann integral. In Measure Theory, there is another integral called Lebesgue integral. It is natural to ask : Why is Lebesgue integral needed?

  1. Riemann integral is partitioned by the range of function f; Lebesgue integral is prtitioned by the domain of function f

It is like two persons are counting their money. What Mr. Riemann does is: add the cashes one by one in order. What Mr. Lebesgue does is: sort the cash, put cashes with the same value together then count the number of $100, number of $50…….. What Lebesgue does seems to be more effective.

  1. Riemann integrable function space is not complete but the sapce for Lebesgue integrable function is complete. i.e, the limitation of a convergence sequence of Riemann integrable functions may not be Riemann integrable. But the limitation of a convergence sequence of Lebesgue integrable is Lebesgue integrable.

  2. Riemann integral requires stronger precondition to interchange the order of limitaion and integration that is: uniform convergence. In Lebesgue, we only need ``Lebesgues dominated convergence theorem (DCT)’’

  3. Let f be a bounded function on a bounded interval [a, b]. Then

You need to show that: upper-Riemann integral=lower-Riemann integral

Proof (to be cont)

  1. If f is Riemann absolute integrabe on an unbounded area \(G\) and on every bounded subarea of \(G\) is Riemann integrable, then f is Lebesgue integrable on \(G\) and the two integrals are coincide.

To show this, you need to use MCT and DCT

For above \(f(x_{1},x_{2},...,x_{n})\), Riemann absolute integrable is equivalent to Riemann intnergable when \(n\geq2\)}

0.3 Examples

\[f(x)=\begin{cases} \begin{array}{cc} 1 & x\in Q\\ 0 & x\in R/Q \end{array}\end{cases}\]

\(f(x)\) Legesbue integrable but not Riemann integrable}

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