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Suppose that $f$ is Riemann integrable. Does there exist $\xi\in[a,b]$ such that $$|f(\xi)|\leq\frac{1}{b-a}\int_{a}^{b}|f(x)| \ \text{d}x?$$ If there exists a number $\xi$ such that $|f(\xi)|\leq|f(x)|$ for all $x\in[a,b]$, then I think that I know how to prove it: Since $f$ is Riemann integrable, $f$ is also Lebesgue integrable. So because $|f(\xi)|1_{[a,b]}$ is a simple function bounded by $f$, it follows by definition of the Lebesgue integral of a non-negative function that $$|f(\xi)|(b-a)=|f(\xi)|\lambda([a,b])\leq\int_{[a,b]}|f| \ \text{d}\lambda=\int_{a}^{b}|f(x)| \ \text{d}x.$$ But how do I prove the general case?

Also, is there a way to avoid lebesgue integrals?

EDIT: Since $f$ is not necessarily continuous, I’m not sure if the MVT can be applied.

Chappers
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Calculix
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2 Answers2

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Since $f$ is Riemann integrable, $g=\lvert f \rvert$ is too. Suppose there is no such $\xi$. Then for every $y \in [a,b]$, we have $$ g(y) > \frac{1}{b-a} \int_a^b g . $$ Integrating over $[a,b]$, dividing by $b-a$, and using positivity of the Riemann integral (that is, that it preserves strict inequalities that hold everywhere) then gives $$ \frac{1}{b-a} \int_a^b g > \frac{1}{b-a} \int_a^b g , $$ a contradiction.

Chappers
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  • (+1) And nice to see you're back. How are you? Have you finished your PhD program? – Mark Viola Mar 25 '20 at 18:06
  • @MarkViola Alas, not yet. Teaching, other research distractions, and now other circumstances, have intervened again! Currently trying to find some interesting results about changing the order of upper and lower integrals, but limited progress so far... – Chappers Mar 25 '20 at 18:23
  • Hmm … I'm interested in hearing more about those preliminary results in changing the order of upper and lower integrals. Where can I read more about that? – Mark Viola Mar 25 '20 at 19:07
  • Bullen and Výborný wrote a paper Arzelà's dominated convergence theorem for the Riemann integral. (Boll. Un. Mat. Ital. A (7) 10 (1996), no. 2, 347–353) in which they show that if $f_n \uparrow f$, $L f_n \to f$, where $L$ is the lower Darboux integral. So I first thought about whether one can obtain information like similar to MCT, FL and DCT for nonintegrable functions using $U,L$. The answer turns out to be very much yes, although it's not in a publishable form yet. [...] – Chappers Mar 26 '20 at 02:58
  • [...] The next question is if Fubini (or similar) has anything equivalent. This paper drew my attention to Fichtenholz's theorem (who knew the Riemann integral had such a nice result!), and I've spent the last few days trying to see if anything can be said about that when some things are not integrable. – Chappers Mar 26 '20 at 02:58
  • (The Riemann version of Fubini's theorem is actually very boring in this setting: it just says $ L_{x,y} f \leq L_x L_y f \leq \genfrac{}{}{0pt}{}{L_x U_y f}{U_x U_y f} \leq U_x U_y f \leq U_{x,y} f$, with the obvious notation.) – Chappers Mar 26 '20 at 03:08
  • Interesting. Is there a prototypical problem or a class of problems in physics that these theorems would equip? – Mark Viola Mar 26 '20 at 15:00
  • Not that I'm know: it seems to me to be about what can be said about integration in impoverished function spaces. There is another way of thinking about it, in terms of what the integral could be, characterising it by betweenness ($m\leq f \leq M \implies m(b-a) \leq \int_a^b f \leq M(b-a)$) and additivity on intervals ($\int_a^b f = \int_a^c f + \int_c^b f$): these two properties are enough to recover the Darboux integral: Darboux integrable functions are precisely those for which any "integral" gives the same answer; Lebesgue comes by replacing intervals by measurable sets in the second. – Chappers Mar 26 '20 at 21:12
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Applying the Lagrange mean value theorem to $$ F(x) = \int_a^x {f(t)dt} $$ shows that there is a $\xi \in (a,b)$ such that $$ f(\xi ) =F'(\xi)=\frac{F(b)-F(a)}{b-a}= \frac{1}{{b - a}}\int_a^b {f(t)dt} . $$ Now $$\left| {f(\xi )} \right| = \frac{1}{{b - a}}\left| {\int_a^b {f(t)dt} } \right| \le \frac{1}{{b - a}}\int_a^b {\left| {f(t)} \right|dt} . $$

Gary
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    The hypotheses don't indicate that $f$ is continuous. Isn't that necessary? – ProfOak Mar 25 '20 at 17:45
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    If its Reimann integrable, then it has a countable number of discontinuities (if that helps the arguement) – QC_QAOA Mar 25 '20 at 17:56
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    @QC_QAOA That is false. The set of discontinuities has measure zero (in fact, this is equivalent to Riemann-integrability for bounded functions), but it need not be countable. – Thorgott Mar 25 '20 at 18:06
  • Well, continuity is not necessary. However, we need that $f$ has an antiderivative. – Gary Mar 25 '20 at 18:15
  • @Thorgott Whoops, you are correct. Misremebered the key detail. My bad – QC_QAOA Mar 25 '20 at 18:29