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The parenthesis in the title comes from the fact that there are essentially six versions of the conclusion in what may be called 2nd Mean Value Th. for Int. - not including special variants like those using Stieltjes integrals and ignoring the different hypotheses (like continuous) on the functions involved. By a symmetric version I mean one with two integrals on the right hand side (the ends of the basic interval $[a,b]$ play then the same role), the others are asymmetric with only one integral on both sides, one can "concentrate" on one end $a$ or on the other one $b$ - this is seen as that one among $a$ and $b$ which occurs as a limit of the integral on the right hand side. Each of the two asymmetric and the symmetric versions exists in a strong and a weak version: a weak version uses as factor(s) of the integral(s) on right side a limit(s) of the monotonic function $d(a+)$ or $d(b-)$ - kind of ignoring a possible discontinuity that it might have at $a$ or $b$ - while strong versions use the values of the function $d(a)$ or $d(b)$ or any number(s) such that could be the value at $a$ or $b$ (instead of the actual ones) without contradicting the monotonicity.

So I would like to see a reference(s) to (relatively) short proof(s) of the statement:

If $f,g$ are real-valued functions defined in the compact interval $[a,b]$ ($a<b$), $g$ being $\ge 0$ and decreasing (not necess. strictly i.e. for $a \le c \le d \le b$ one has $g(c) \ge g(d)$ ) and $f$ continuous (may-be not with bounded variation), there exists a $c$ such that $a \le c \le b$ for which $$\int _a^b f(x)g(x)dx = g(a) \int_a^c f(x)dx$$ (some authors mention that $c$ can be taken in the open interval $]a,b[$ under some extra hyp. but I don't need that)

Important: g must be allowed to not be continuous

Motivation: I need this for a proof of a result in the Fourier series (nearly the unique place for using a version of the 2nd MVT for I.) which I currently study in Y.Katznelson's book - he seems to use this implicitly without even mentioning the name of the result used, but I found in old notes from a course on F. series I attended ~ 50 years ago the name of the theorem ... and then in Wikipedia what this could mean, but no proof. In all pseudo-duplicates to be found in this website, none satisfies me. Some use Stieltjes integrals like W.Rudin in his Principles of Mathematical Analysis where this subject occupies ~ 2 to 3 pages but some extra work would be needed to get the above statement and it would not cover the case where $f$ has not bounded variation. I have found a proof but it is 11 pages (which I might reduce to ~ 5 to 7 pages by eliminating parts I don't need, still a bit long) - I guess it is correct: see https://my-calculus-web.web.app/Fourier%20Series/Second%20Mean%20Value%20Theorem%20for%20Integrals.pdf

BTW a symmetric alternative (like the one I link to above) might be OK because one can usually deduce from it easily an asymmetric version.

Ulysse Keller
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    If $g(x)$ isn't continuous, it just won't work, the required intermediate value of $g$ might just not exist. – vonbrand May 26 '24 at 18:50
  • This is Bonnet's mean value theorem. The usual hypotheses are that $g$ be monotone (not necessarily continuous) and $f$ be continuous. The continuity of $f$ can even be relaxed such that $f$ is Riemann integrable. – RRL May 26 '24 at 20:27
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    @vonbrand with this theorem and similar versions intuition can make us think that way, but RRL must be right, and in the (long) proof that I linked to $f$ may even be just Lebesgue integrable ... – Ulysse Keller May 27 '24 at 21:33

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There are relatively easy proofs under the weaker assumption that $f$ is Riemann integrable (rather than continuous) and $g$ is nonnegative and decreasing.

One approach is to use Riemann sums. It is straightforward but perhaps less elegant. Reference: A Radical Approach to Real Analysis by Bressoud.

Another approach, that brings in the Riemann-Stieltjes integral as an intermediate step, is to introduce the function $F(x) = \int_a^x f(t) \,dt$. Note that $F$ is continuous given that $f$ Riemann integrable and, hence, bounded.

The key step relies on the fact that with both $f$ and $g$ Riemann integrable we have

$$\int_a^b f(x) g(x) \, dx = \int_a^b g \, dF,$$

and then using integration by parts it follows that

$$\tag{*}\int_a^b f(x) g(x) \, dx= g(b)F(b) - g(a)\underbrace{F(a)}_{= 0}- \int_a^bF\, dg$$

Since $F$ is continuous there exists, by the first mean value theorem for integrals$, c \in (a,b)$ such that

$$\int_a^bF(x)\, dg = F(c)[g(b)-g(a)]$$

Substituting into (*) we get,

$$\int_a^b f(x) g(x) \, dx= g(b)F(b) - F(c)[g(b)-g(a)]=g(a)\int_a^cf(x) \, dx +g(b)\int_c^bf(x) \, dx $$

Apply this result to the decreasing function $\hat{g}(x) = \begin{cases}g(x), & a\leqslant x < b\\0,&x = b \end{cases}$ to obtain

$$\int_a^b f(x) g(x) \, dx=g(a)\int_a^cf(x) \, dx $$

RRL
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  • The fact that $\int_a^b f(x) g(x) , dx = \int_a^b g , dF$ can be proved directly using Rieman-Stieltjes sums. – RRL May 26 '24 at 20:37
  • Good, but I must look how to insert one of the two proofs into my context. That is nearly always the problem with proofs obtained in such a way: there might be prerequisites that are not the same for the one who asks and the one who answers. For instance I know about Stieltjes integrals but not a theorem of Integration by parts for them - the one you use looks "probably correct" for one who knows any theorem of this kind, but that isn't a proof ... I need some time to analyze this. – Ulysse Keller May 27 '24 at 21:14
  • @UlysseKeller: I agree. A self-contained proof requires a justification of integration by parts for the Riemann-Stieltjes integral. It is quite easy and involves a straightforward manipulation of Riemann-Stieltjes sums, which you can find in part (2) of this answer – RRL May 28 '24 at 20:17
  • There are 3 results to combine: a) the one about which you added a comment (a no-name as far as I know), b) Stieltjes-integration by parts, c) first MVT for int. - for a) I have found something / for b) after trying something myself I understand now the essentials of the proof by Rudin in the book I mentioned / for c) one finds what is necessary in Rudin as well as in Bressoud (both have essentially the same argument) -> the rest being simple, I only have to clear something for b) concerning limit processes to be done – Ulysse Keller May 28 '24 at 22:56
  • I like to consider integration processes as the effect of linear continuous maps defined in a normed vector space of functions and scalar values - real here, sometimes complex (ideas can be extended to integration of vector-valued functions); in Riemann-style integrals (incl. Stieltjes) continuity is w.r.t. the supremum-norm (leading to the topology of uniform convergence); in Lebesgue-style integrals continuity is w.r.t. $L^1$ norm. The main idea is to start with a definition on a dense subspace, and then to extend by continuity ... (continued) – Ulysse Keller May 28 '24 at 23:09
  • (Cont. of last comment) in the case of Riemann style, one begins with characteristic functions of sub-intervals of [a,b](not a vector space) where - a discovery of mine which I have not yet seen in math. lit. - taking only intervals [a,u] (u in [a,b] ) and [a,u[ (u in ]a,b] ), these char. funct. are linearly independent, so you are free to define the linear map on them (using the g in "dg") and they generate the subspace of so-called step or staircase functions. The closure of this is the space of regulated functions (see wikipedia) so one gets integrals of such functions. I use this for a) – Ulysse Keller May 28 '24 at 23:38