For an interval $[a,b]$, let ${\mathscr P}(a,b)$ be the set of all partitions $P=\lbrace x_0,x_1,\ldots,x_n\rbrace$ of $[a,b]$, where $a=x_0<x_1<\cdots <x_n=b$.
For a bounded function $f:[a,b]\rightarrow {\mathbb R}$ and $P=\lbrace x_0,x_1,\ldots,x_n\rbrace$, let
$$L(f,P)=\sum_{i=1}^n m_i(x_i-x_{i-1}),$$
where $m_i=\inf\lbrace f(x):x\in [x_{i-1},x_i]\rbrace$
and
$$U(f,P)=\sum_{i=1}^n M_i(x_i-x_{i-1}),$$
where $M_i=\sup\lbrace f(x):x\in [x_{i-1},x_i]\rbrace$
Many Real Analysis books define $f$ to be Riemann integrable if
$$ L(f)=\sup\lbrace L(f,P): P\in {\mathscr P}(a,b)\rbrace=\inf\lbrace U(f,P): P\in {\mathscr P}(a,b)\rbrace=U(f).$$
Now let $|P|$ be the mesh of the partition $P$, that is, $|P|=\max\lbrace x_i-x_{i-1}:1\leq i \leq n\rbrace$.
A tag for $P=\lbrace x_0,x_1,\ldots,x_n\rbrace$ is a sequence $t=\lbrace t_1,t_2,\ldots,t_n\rbrace$ such that $t_i\in [x_{i-1},x_i]$ for $1\leq i \leq n$.
We denote by ${\mathscr P}^\ast(a,b)$ the set of all pairs $(P,t)$ where $P\in {\mathscr P}(a,b)$ and $t$ is a tag for $P$.
If $(P,t)\in {\mathscr P}^\ast(a,b)$, let
$$S(f,P,t)=\sum_{i=1}^n f(t_i)(x_i-x_{i-1}).$$
Other Real Analysis books define $f$ to be Riemann integrable if there is a real number $I$ such that
$$(\forall \epsilon >0)(\exists \delta>0)(\forall (P,t)\in {\mathscr P}^*(a,b))(|P|<\delta
\Rightarrow |S(f,P,t)-I|<\epsilon).$$
Question: what is the simplest way to see that the two definitions are equivalent? I am looking for a proof that does not involve too many concepts other than these, and the basic facts that $L(f)\leq U(f)$ is always true, and that if $P\subset Q$, then
$L(f,P)\leq L(f,Q)$ and $U(f,P)\geq U(f,Q)$. I am especially interested in proving that if $f$ is Riemann integrable according to the first definition, then it is Riemann integrable according to the second definition.
Best Answer
Here's an outline of the most basic proof I know. Let $\epsilon>0$ be given. Since $f$ is Riemann integrable, there is $M>0$ so that $|f|\le M$. Fix a partition $\mathscr P'=\{a=x'_0, x'_1,\dots, x'_r=b\}$ of $[a,b]$ so that $U(f,P')-L(f,P')\epsilon/2$. Now, take an arbitrary partition $\mathscr P=\{a=x_0, x_1,\dots,x_n=b\}$ with mesh $<\delta$. We divide $\mathscr P$ into the set of "good" intervals and the set of "bad" intervals: Each good interval is contained in some interval of $\mathscr P'$; each bad interval fails to be contained in some interval of $\mathscr P'$. We denote by $\mathscr P'_{\text{good}}$, resp. $\mathscr P_{\text{good}}, the union of the good intervals of the appropriate partition; similarly for bad.
Now $$L(f,\mathscr P'_{\text{good}})\le S(f,\mathscr P_{\text{good}},t)\le U(f,\mathscr P'_{\text{good}}).$$ What about the bad intervals? Each one must contain one of the $x'_i$ in its interior, and so there are at most $r-1$ such intervals; their total length, then, is less than $(r-1)\delta$. Thus, $$-M(r-1)\delta<S(f,\mathscr P_{\text{bad}},t)<M(r-1)\delta.$$ If we require $\delta<\dfrac{\epsilon}{4M(r-1)}$, then $$-\frac{\epsilon}4<S(f,\mathscr P_{\text{bad}},t)<\frac{\epsilon}4.$$
Letting $I_{\text{good}}$, resp. $I_{\text{bad}}$, denote the integral of $f$ over the good, resp. bad, intervals of $\mathscr P$, we have $$\left|S(f,\mathscr P_{\text{good}},t)-I_{\text{good}}\right| \le U(f,\mathscr P'_{\text{good}}) - L(f,\mathscr P'_{\text{good}}) < \frac{\epsilon}2$$ and $$\left|S(f,\mathscr P_{\text{bad}},t)-I_{\text{bad}}\right| < \frac{\epsilon}2;$$ adding the two gives the desired result.