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Theorem :

If $f$ is Riemann integrable on $[a,b]$, Then so is $f^2$.

There are many proofs for this theorem. But i don't want the proofs which use $U(f,P)$ and $L(f,P)$. My book gives other definitions :

$S(f,P^t)=\sum_{i=1}^n f(t_i)(x_i-x_{i-1})$

$\omega(f,[a,b])=\sup\{f(x):x\in[a,b]\}-\inf\{f(x):x\in[a,b]\}$

We say $f:[a,b] \to \mathbb R$ is Riemann integrable on $[a,b]$, if there exists $L$ such that :

For each $\epsilon \gt 0$ there exists $\delta \gt 0$ such that for each labeled partition $P^t$ such that $||P^t|| \lt \delta$ , We have :
$|S(f,P^t)-L|\lt \epsilon$

This definition is not related to $U(f,P)$ and $L(f,P)$ and The question wants us to proof the above theorem with this definition and the properties of $\omega(f,[a,b])$. I think this property helps :

$\omega(f^2,[a,b]) \le 2M\omega(f,[a,b])$ ( $M$ is the bound of $f$ )

My problem is that i can't rewrite this definition in a way that $f^2$ becomes Riemann integrable.

Edit :

We have two related theorems which i think may help :

  1. A bounded function $f$ is Riemann integrable on $[a,b]$ if and only if for each $\epsilon \gt 0$ there exists $\delta \gt 0$ such that for each two partitionings $P_1^t$ and $P_2^t$ such that their norms are less than $\delta$ , $|S(f,P_1^t)-S(f,P_2^t)| \lt \epsilon$

  2. Assume that $f$ is a bounded function defined on $[a,b]$. Assume that $P_1=\{z_i:0\le i\le n\}$ is a partitioning for $[a,b]$ and $P_2$ is an elegant partitioning for $P_1$ ( Meaning it has more points ) If $P_1^t$ and $P_2^t$ are two labeled partitionings for $[a,b]$ derived from $P_1$,$P_2$ , Then :

$|S(f,P_1^t)-S(f,P_2^t)| \le \sum_{i=1}^n \omega(f,[z_{i-1},z_i])(z_i-z_{i-1})$

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    Which role does $\omega$ play in your definition ? – 2017-01-04
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    @A.MONNET it doesn't play "a role". But, the question wants to use it somehow... I read this proof and came up with the property that i wrote about $\omega$ : http://www.math.lsa.umich.edu/~canary/HW8.pdf – 2017-01-04
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    As written, your definition is a bit ambiguous, because you don't say what $t_i$ is. – 2017-01-04
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    Would it be possible to prove a theorem for the composition $g\circ f$ ? where $g$ is continuous and $f$ is Riemann integrable. Then you can conclude by taking $g(x) = x^2.$ – 2017-01-04
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    @MartinArgerami $t_i$ is just a point between $x_{i-1}$ and $x_i$ such that $x_{i-1}$ and $x_i$ are two points of $P$. – 2017-01-04
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    I know, but I was talking about how everything is written. I guess you mean that $P^t$ is a partition $\{x_j\}$ together with points $t=\{t_j\}$ and $x_j – 2017-01-04
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    @A.MONNET we have a cauchy theorem for riemann integration... i'll write it ... wait... i think that's useful ... – 2017-01-04
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    @MartinArgerami yes, you're right – 2017-01-04
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    Using the Cauchy criterion for the Riemann integral, it should be easy to adapt the more "usual" proof once you observe that $L(P,f)\le S(P^{t},f)\le U(P,f)$. – 2017-01-04
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    @chilangoincomprendido I can't use the upper and lower summations !!! My definition of integration is different in the book ! – 2017-01-04
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    But you can easily adapt the proof using $\omega(f^2,[a,b]) \le 2M\omega(f,[a,b])$. – 2017-01-04
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    @chilangoincomprendido it isn't so easy in my mind ... if you know how to adapt that, please write an answer :) – 2017-01-04
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    hint: Applying 2) to $f^2$ you have $|S(f^2,P_1^t)-S(f^2,P_2^t)| \le \sum_{i=1}^n \omega(f^2,[z_{i-1},z_i])(z_i-z_{i-1})$ and this is $\le 2M\sum_{i=1}^n \omega(f,[z_{i-1},z_i])(z_i-z_{i-1}).$ – 2017-01-05
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    Note also that this is not really a different proof, because if $P=\left \{ a,x_1,\cdots, x_{n-1},b \right \}$ and if $M_i$ and $m_i$ are the max/min, resp. of $f$ on $[x_i,x_{i+1}]$ then $M_i(f)-m_i(f)=\omega(f,[x_i,x_{i+1}])$, so you are, in fact, making use of upper and lower sums. – 2017-01-05
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    Your definition is the one originally given by Riemann and it is equivalent to the more convenient definition using upper lower Darboux sums. But if you insist on using Riemann definition then I think Andrew D. Hwang's answer gives the simplest way to handle this question. – 2017-01-05
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    @ParamanandSingh can you please explain how should i continue his answer to reach the proof ? – 2017-01-05
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    Use point 2 of your question to get $$|S(f^{2},P1)-S(f^{2},P2)|\leq 2M\sum\omega(f,[z_{i-1},z_{i}])(z_{i}-z_{i-1})$$ and then use point 1 of Andrew's answer to place RHS above by $2M|S(f,P1)-S(f,P2)|+2M\epsilon$ and by point 1 of your question this is less than $4M\epsilon$. Thus by Cauchy's criterion $f^{2}$ is Riemann integrable. – 2017-01-05

1 Answers 1

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$\newcommand{\eps}{\varepsilon}$Hints using your book's definitions:

  1. For every $\eps > 0$ and for every partition $P = (z_{i})_{i=0}^{n}$ of $[a, b]$, there exist labelings $P_{1}^{t}$ and $P_{2}^{t}$ such that $$ \sum_{i=1}^{n} \omega(f, [z_{i-1}, z_{i}])(z_{i} - z_{i-1}) \leq |S(f, P_{1}^{t}) - S(f, P_{2}^{t})| + \eps. \tag{*} $$ Proof: For each subinterval $I = [z_{i-1}, z_{i}]$, pick the label $t_{i}$ of $P_{1}^{t}$ so that $$ 0 \leq \sup_{x \in I} f(x) - f(t_{i}) < \frac{\eps}{2(b-a)}, $$ and pick the label $t'_{i}$ of $P_{2}^{t}$ so that $$ 0 \leq f(t'_{i}) - \inf_{x \in I} f(x) < \frac{\eps}{2(b-a)}. $$ Rearranging gives $$ \sup_{x \in I} f(x) - \inf_{x \in I} f(x) < f(t_{i}) - f(t'_{i}) + \frac{\eps}{b - a}. $$ Summing over $i$ gives inequality (*).

  2. For each interval $I$ contained in $[a, b]$, you have $\omega(f^{2}, I) \leq 2M\, \omega(f, I)$.

Can you take it from here?

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    +1 for the first result which you have given. Nice approach. – 2017-01-05
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    @AndrewD.Hwang excuse me ... i'm soooo sorry ... but i still don't get what happens next ... would you please explain it ? – 2017-01-05
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    @Arman: Did Paramanand's sketch (in the main comments) help? Note that: 1. You _have_ to use the "Cauchy criterion" because there's no way to express the integral of $f^{2}$ as a function of $L$, the integral of $f$. 2. Thanks to the definition of integrability, it's essential to dominate (i.e., find in upper bound for) $\Delta(f, P) := U(f, P) - L(f, P)$ by a difference of Riemann sums for $f$. Unfortunately the difference between _arbitrary_ Riemann sums is bounded above by $\Delta(f, P)$. The only recourse is to show "a difference of Riemann sums plus $\eps$ exceeds $\Delta(f, P)$". – 2017-01-05
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    @AndrewD.Hwang yes... and thanks for your detailed answer and for the attention to my question :) – 2017-01-05