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I've been reviewing proofs for a couple of calculus theorems and as I was trying to recall the proof of the Riemann Sum Theorem which uses Lower Sums and Upper Sums I came up with an idea to prove it using The Mean Value Theorem for Integrals.


I just wanted to know if my proof is good since I'm not sure if I've done everything properly. If you've got any hints for me, I'd be delighted.

My manipulations in the 3. step might be based a "little bit" too much (just a little bit I hope!) on my "intuitions". Excuse me, if this violation of Mathematical principals send shiver down your spine today.


Proof

  1. Just to make sure there is no misunderstanding - this is what I call the "Riemann Sum Theorem":
    Assuming that:
    1. $f: [a;b]\to\mathbb{R}$.
    2. ${P_n}$ is a sequence of partitions of $[a;b]$ such that $|P_n|\to0$ where $|P_n|$ is the length of the longest partition.
    3. $f$ continuous.
    4. $S_n$ is a Riemann Sum of $f$ and $P_n$.
    Then $S_n(f, P_n) \to \int_{[a;b]}{f}$.
  2. We know that $S_n(f, P_n)=\sum_{j=1}^nf(y_j)(x_{j}-x_{j-1})$ where $y_j\in[x_{j-1};x_j]$, $j=\{1,\dots,n\}$ and $P_n=\{x_0,\dots,x_n\}$.
  3. $S_n(f, P_n) = f(y_1)(x_1-x_0)+\dots+f(y_n)(x_n-x_{n-1})$ and it starts to resemble something we can apply the Mean Values Theorem to. Especially, that $$ \forall_{k \in [1;n]}\forall_{\epsilon>0}\exists_{\delta>0}\forall_{u,v \in [x_{k-1};x_k]}\left(|u-v|<\delta \implies |f(u)-f(v)|<\epsilon\right) $$
  4. So, as $n \to \infty$: $$ \begin{cases} f(y_1)(x_1-x_0) \to \int_{[x_0;x_1]}f\\ f(y_2)(x_2-x_1) \to \int_{[x_1;x_2]}f\\ \vdots\\ f(y_n)(x_n-x_{n-1}) \to \int_{[x_{n-1};x_n]}f\\ \end{cases} $$ which "sums up" to $$ \int_{[x_0;x_n]}f = \int_{[a;b]}f $$
  5. $\square\dots$ is it?

Note: Sorry if I use unpopular and weird names for theorems. I don't study Mathematics in English and it is difficult to find the exact translation from time to time. Feel free to edit my posts if necessary.

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    $\begingroup$ What you are calling the "Riemann sum theorem" is actually the definition of Riemann integrals, so there is nothing to prove. On top of which, you may not assume any theorems about an undefined object, so if you have not defined Riemann integrals then you may not assume the Mean Value Theorem for Integrals. $\endgroup$ – Lee Mosher Sep 2 '15 at 16:46
  • $\begingroup$ Thank you @LeeMosher! I should have defined Riemann Integrals as well since I am defining Riemann Integral as the value of $\int_{[a;b]}f$ when $\int_{[a;b]}f = \overline{\int_{[a;b]}f} = \underline{\int_{[a;b]}f}$ (I believe these are called Upper and Lower Darboux Integral). Therefore I do not treat Riemann Sum Theorem as a definition of Riemann Integrals. For sure, there is something wrong with my reasoning but I haven't understood yet what it might be. $\endgroup$ – Mateusz Piotrowski Sep 2 '15 at 18:04
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It is possible.

First of all note that $\int_a^b f$ is a Riemann sum w.r.t. any partition of $[a,b]$ as a consequence of the additivity on intervals and the MVT for integrals.

Hence, if $P$ is a tagged partition of $[a,b]$, then $$\sum_{k=1}^n f(\xi_k)(x_k-x_{k-1})-\int_a^b f=\sum_{k=1}^n [f(\xi_k)-f(y_k)](x_k-x_{k-1})$$ So, if $N$ is such that $\|P_n\|<\delta$ for every $n>N$, then $$\left|S(f,P_n)-\int_a^b f \right|<\varepsilon(b-a)$$ for every $n>N$.

Of course $\varepsilon$ and $\delta$ are those appearing in your post.

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