</head>

In ``Group Theory’’, the basic operation is \("\times"\), we define the kind of mapping which keeps the operation as : homomorphic mapping. It is a key mapping in Group Theory.

In Mathematical Analysis, the foodstone is

**limit**and the continuous functions are that keep**limit operation**, i.e.

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

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

- Different definition of Continuous Functions:

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

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

- 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.

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

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?

- 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.

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**.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)’’Let f be a bounded function on a bounded interval [a, b]. Then

- f is Riemann integrable on [a,b] iff f is continuous i.e (m) on [a, b]

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

Proof (to be cont)

- f is Lebesgue integrable on [a, b] and the Lebesgue integral \(\int_{[a,b]}fdm\) equals the Riemann integral \(\oint_{[a,b]}f\), i.e., the two integrals coincide.

- 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 \(n\geq2\)}

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

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

</html>