First definition of Riemann integrable function. Let $f:[a,b] \to \mathbb{R}$ be a bounded function and $P=\{x_0,x_1,\dots, x_n\}$ partition of $[a,b]$. Define $U(P,f):=\sum \limits_{i=1}^{n}M_i\Delta x_i$ and $L(P,f):=\sum \limits_{i=1}^{n}m_i\Delta x_i$ where $M_i=\sup\limits_{[x_{i-1},x_i]} f(x),\quad m_i=\inf\limits_{[x_{i-1},x_i]} f(x), \Delta x_i=x_i-x_{i-1}.$ Let $\inf \limits_{P}U(P,f)=I^*$ and $\sup \limits_{P}L(P,f)=I_*$. If $I^*=I_*=I$ then we called $f(x)$ is Riemann integrable function on $[a,b]$ with integral $I$.
Second definition of Riemann integrable function. Let $P=\{x_0,x_1,\dots, x_n\}$ is a partition of $[a,b]$ with $\xi_i\in [x_{i-1},x_i]$ and we define Riemann-integral sum $\sigma(P):=\sum \limits_{i=1}^{n}f(\xi_i)\Delta x_i$ and $\lVert P\rVert=\max\limits_{i}\Delta x_i$. If the following limit $\lim \limits_{\lVert P\rVert\to 0}\sigma(P)$ exists and has value $L$ we say that $f(x)$ is Riemann integrable function on $[a,b]$ with integral $L$.
The first definition is from Rudin's PMA book but in other books I met the second definition. But I can't prove the equivalence of these definitions for a couple days. Can anyone show to me a strict and rigorous proof? I would be very grateful for your help!
P.S. Happy New Year! 🙂
Best Answer
That the first condition implies the first is immediate, since (using your notation) you always have $m_i \le f(\xi_i) \le M_i$, so the sums in the second definition are caught between the $L$ und $U$ sums.
Edit in response to a comment an additional explanation is necessary here. For this direction it suffices to show that $I^* = lim_{||P||\rightarrow 0} L(f,P)$ and $I_* = lim_{||P||\rightarrow 0} U(f,P)$ Since both parts are similar it suffices to show, e.g., the first equality.
First it is easy to see that for partitions $P\subset P^\prime$ we have $L(f,P)\le L(f,P^\prime)$. A remaining hurdle is that for two partitions we do not necessarily know that one is a subset of the other one. This is resolved by looking at common refinements:
Assume $P$ satisfies $L(f,P) > I^* - \varepsilon$ and $Q$ is an arbitrary partition. We need to show that then there is a refinement $Q^\prime$ of $Q$ such that $L(f,Q^\prime)\ge L(f,P)$ (and, consequently, $L(f,Q^\prime)>I^*-\varepsilon$).
For $Q^\prime$ one can choose the common refinement $R$: if $P=\{x_1,\ldots x_n \}$ and $Q=\{y_1,\ldots y_m \}$ then we just let $R = P\cup Q$. Since this is a refinement of both $P$ and $Q$ we have both $L(f,R)\ge L(f,P)$ as well as $L(f,R)\ge L(f,Q)$
Second edit: the original version was not correct:
For the other direction it suffices to show that if the function is integrable in the sense of the second definition then both $I_*$ and $I^*$ agree with the of the sums from the second definition. Since the reasoning is the same in both cases I'll just look at $I_*$.
So fix $\varepsilon >0$ and a given partition $P$ such that $$|L - \sum_{i=1}^n f(\xi_i)\Delta x_i |< \varepsilon$$ if only the partition is fine enough.
Choose such a partition $P=\{x_0,\dots x_n\}$ and to $[x_{i-1},x_{i}]$ choose $\eta_i\in[x_{i-i},x_{i}]$ such that for $m_i:=\inf \{ f(x):x\in [x_{i-1},x_i]\} $ we have $$0\le f(\eta_i)-m_i\le \frac{\varepsilon}{2n}$$
Then \begin{eqnarray} | L -\sum_{i=1}^n m_i \Delta x_i| & = & |L- \sum_{i=1}^n f(\eta_i)\Delta x_i + \sum_{i=1}^n f(\eta_i)\Delta x_i -\sum_{i=1}^n m_i\Delta x_i| \\ &\le & |L- \sum_{i=1}^n f(\eta_i)\Delta x_i| + \sum_{i=1}^n | f(\eta_i) - m_i|\Delta x_i \\ & < & \frac{\varepsilon}{2} + \sum_{i=1}^n \frac{\varepsilon}{2n}=\varepsilon \end{eqnarray}
If you 'see' that $0 <L -I_*< L -\sum_{i}m_i \Delta x_i$ then you are done here, otherwise it follows easily from the last estimate that the $\sum_i m_i \Delta x_i$ are, for any partition which is fine enough, $\varepsilon $ close to the fixed real number $L$, which of course implies that the $\sup$ over these sums exists and equals $L$ (here you need to use again the fact that you will approach the $\sup$, if it exists, if the width of the partitions goes to $0$).