Riemann Integral

Intro

There exists systems modeled by equations of the form x(ti+1)=x(ti)+f(ti,x(ti))(ti+1−ti)    0=t0<t1<⋯<tN=1(1)\tag{1} x(t_{i+1})=x(t_{i})+f(t_{i},x(t_{i}))(t_{i+1}-t_{i}) \ \ \ \ 0=t_{0}<t_{1}<\cdots<t_{N}=1 “In the limit” as (t0,…,tN)(t_{0},\dots,t_{N}) of [0,1][0,1] gets finer (i.e. max⁡i∣ti+1−ti∣→0\max_{i}\mid t_{i+1}-t_{i}\mid\to 0) leads to an integral equation of the form x(t)=x(0)+∫0tf(s,x(s)) dsx(t)=x(0)+\int \limits_{0}^tf(s,x(s)) \, ds where the integral is the Riemann Integral.

Problems with The Riemann Integral

Let fn:[0,1]→Rf_{n}:[0,1]\to \mathbb{R} be a sequence of Riemann-integrable functions on [0,1][0,1], there are two issues:

  1. Assume ∃f:[0,1]→R\exists f:[0,1]\to \mathbb{R} such that fn(x)→f(x)f_{n}(x)\to f(x) as n→∞n\to \infty, ∀x∈[0,1]\forall x \in[0,1] (i.e. (fn)n∈N(f_{n})_{n\in\mathbb{N}} converges pointwise on [0,1][0,1] to ff) then, it does not follow that ff is Riemann-integrable
  2. Even if ff happens to be Riemann-integrable, it does not follow that lim⁡n→∞∫01fn(x) dx=∫01f(x) dx\lim_{ n \to \infty } \int\limits_{0}^{1}f_{n}(x) \, dx =\int\limits _{0}^{1}f(x) \, dx i.e., the limit and Riemann integral cannot be always interchanged.

Definition (Riemann integral)

A function f:[0,1]→Rf:[0,1]\to \mathbb{R} is Riemann integrable if ∀ϵ>0,∃δ>0\forall\epsilon>0, \exists\delta>0 such that for each subdivision π=(t0,…,tN(π))\pi=(t_{0},\dots,t_{N(\pi)}) of [0,1][0,1] we have that if the subdivision is fine enough then δ(π)<δ  ⟹  ∣Sf,πu−Sf,πl∣<ϵ\delta(\pi)<\delta \implies|S_{f,\pi}^{u}-S_{f,\pi}^{\mathscr{l}}|<\epsilon

Definition (Darboux sums)

Let f:[0,1]→Rf:[0,1]\to \mathbb{R} and let π=(t0,…,tN)\pi=(t_{0},\dots,t_{N}) be a subdivision of [0,1][0,1]. For each i∈{0,…,N−1}i\in\{0,\dots,N-1\} let mi=inf⁡{f(t):t∈[ti,ti+1]}Mi=sup⁡{f(t):t∈[ti,ti+1]} \begin{align*}m_{i}&=\inf\{f(t):t\in[t_{i},t_{i+1}]\}\\ M_{i}&=\sup\{f(t):t\in[t_{i},t_{i+1}]\}\end{align*}We define the Lower Darboux sum of ff with respect to π\pi by Sf,πl=∑i=1N−1mi(ti+1−ti)S_{f,\pi}^{\mathscr{l}}=\sum_{i=1}^{N-1}m_{i}(t_{i+1}-t_{i}) and the Upper Darboux sum of ff with respect to π\pi by Sf,πu=∑i=1N−1Mi(ti+1−ti)S_{f,\pi}^{\mathscr{u}}=\sum_{i=1}^{N-1}M_{i}(t_{i+1}-t_{i})

Proposition

The Riemann Integral exists for all piecewise Continuous functions.

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