Proof of Second Mean Value Theorem for Integrals

Question

This is from wikipedia on MVT.

If $G: [a, b] \to \mathbb{R}$ is a positive monotonically decreasing function and $\phi : [a, b]\to \mathbb{R}$ is an integrable function, then there exists a number $x \in (a, b]$ such that $$ \int^b_a G(t)\phi(t)\,dt = G(a^+)\int^x_a\phi(t)\,dt$$ Here $G(a^+)$ stands for $\lim_{x\rightarrow a^+} G(x)$, the existence of which follows from the conditions. Note that it is essential that the interval $(a, b]$ contains $b$.

• I can prove it for the case when $\phi\geq0$. Let $f(x)=\int^x_a\phi(t)\,dt$, then $0\leq f(x)\leq f(b)$. I also have $0\leq\int^b_a ()\phi(t)\,dt\leq G(a^+)f(b)$. As $f(x)$ is continuous, dividing the last equation by $G(a^+)$ yields our desired relation.
• If $\phi$ is integrable, I tried splitting into positive and negative parts $\phi_+$ and $\phi_-$. But did not proceed far? Any hints?

Answer 1

First assume $G$ is a positive decreasing staircase function, say $G= \sum_{i=1}^n f_i \cdot \chi_{]x_{i-1}, x_i[}$, where $a= x_0 < x_1 < \cdots<x_n=b$, $f_i$ are positive real numbers and $\chi_{]x_{i-1}, x_i[}$ denotes the indicator function of the interval $]x_{i-1}, x_i [$. Let $H: x \in [a,b] \mapsto \int_a^x \phi $. The function H is (absolutely) continuous. By the intermediate value theorem it suffices to show that $$ m G(a^+) \leq \int_a^b G\phi \leq M G(a^+)$$ where $ m = \min H$ and $M = \max H$. Notice that $f_1 = G(a^+)$. Now, \begin{align}\int_a^b G\phi &= \sum_{i=1}^n \int_{x_{i-1}}^{x_{I}}f_i\phi \\ & = \sum_{i=1}^n f_i (H(x_{i}) - H(x_{i-1})) \\ & = \sum_{i=1}^n f_i H(x_i) - \sum_{i=1}^n f_i H(x_{i-1})\\ & = \sum_{i=1}^{n-1}\underbrace{(f_i - f_{i+1})}_{\geq 0}H(x_i) + f_nH(b)\end{align} This yields $$ m\left(\sum_{i=1}^{n-1}(f_i -f_{i+1}) + f_n \right) \leq \int_a^b G\phi \leq M\left(\sum_{i=1}^{n-1}(f_i -f_{i+1}) + f_n \right) $$ and thus we get the inequality which was desired, which proves the result.

Now you can easily approximate $G$ as a limit of staircase positive decreasing functions to conclude.

Answer 2

For a different approach, the function $\varPhi: x \mapsto \int_a^x \varphi(t) \, dt$ is continuous and since $G$ is monotone we have that $\varPhi$ is Riemann-Stieltjes integrable with respect to $G$. From integration by parts, it follows that $G$ is integrable with respect to $\varPhi$ and

$$\int_a^b G \, d\varPhi= G(b)\varPhi(b) - G(a)\varPhi(a) - \int_a^b \varPhi \, dG = G(b)\varPhi(b) - \int_a^b \varPhi \, dG.$$

It also follows from a general property of the Riemann-Stieltjes integral that if $G$ and $\varphi$ are Riemann integrable, which is the case here, then

$$\int_a^bG(t) \varphi(t) \, dt = \int_a^b G \, d\varPhi = G(b)\varPhi(b) - \int_a^b \varPhi \, dG.$$

Since $\varPhi$ is continuous, by the first mean value theorem for integrals there exists $x \in [a,b]$ such that

$$\int_a^b \varPhi \, dG = \varPhi(x)\int_a^b dG = \varPhi(x)[G(b) - G(a)].$$

Hence,

$$\begin{align}\int_a^bG(t) \varphi(t) \, dt &= G(b)\varPhi(b) - \varPhi(x)[G(b) - G(a)] \\ &= G(a) \varPhi(x) + G(b)[\varPhi(b) - \varPhi(x)] \\ &= G(a) \int_a^x \varphi(t) dt + G(b) \int_x^b \varphi(t) \, dt \end{align}$$

Since we can obtain the same result by replacing $G$ with $\hat{G}$ where $\hat{G}(t) = G(t) $ on $[a,b)$ and $\hat{G}(b) = 0$ we obtain

$$\begin{align}\int_a^bG(t) \varphi(t) \, dt &= \int_a^b\hat{G}(t) \varphi(t) \, dt \\ &= \hat{G}(a) \int_a^x \varphi(t) dt + \hat{G}(b) \int_x^b \varphi(t) \, dt \\ &= G(a) \int_a^x \varphi(t) dt \end{align} $$

Note that there is no reason why we must restrict $x \in (a,b]$. This would become evident by working through the proof of the first mean value theorem for integrals where no such restriction applies. If $\varPhi$ is continuous, attaining minimum and maximum values at $\xi_1$ and $\xi_2,$ and it happens that $\varPhi(\xi_1) < \varPhi(a) < \varPhi(\xi_2)$ then $\varPhi(a) = \varPhi(x)$ for at least one other point in $(a,b)$. It is possible that this is a point where $\varPhi(x) = \int_a^b \varPhi \, dG / [G(b) - G(a)]$.

Answer 3

Here is a slightly more general result that extends the OP statement for Lebesgue integrable functions:

Let $m$ denote the Lebesgue measure on the real line.

Proposition: Suppose $g$ is a monotone right-continuous with left limits function defined on an interval $[a,b]$, and $f$ is a real-valued function in $L_1([a,b],m)$. There is a point $x_0\in[a, b]$ such that \begin{align}\int^b_a f(t)g(t-)\,dt= g(a)\int^{x_0}_af(t)\,dt+g(b)\int^b_{x_0}f(t)\,dt\tag{0}\label{zero} \end{align}

• If $g\geq0$ and monotone nondecreasing, then a point $x_0\in[a,b]$ can be chosen so that \begin{align}\int^b_a f(t)g(t-)\,dt= g(b)\int^b_{x_0}f(t)\,dt\tag{1}\label{one} \end{align}
• If $g\geq0$ and monotone non increasing, then a point $x_0\in[a,b]$ can be chosen so that \begin{align}\int^b_a f(t)g(t-)\,dt= g(a)\int^{x_0}_af(t)\,dt\tag{2}\label{two} \end{align}

Comment The integral $\int^b_af(t)g(t-)\,dt$ can be change to $\int^b_b f(t)g(t)\,dt$ since $g(t)=g(t-)$ $m$-a.s. However, it is convenient to keep the former expression to make the application of Lebesgue integration by parts clearer.

Proof: Suppose $g$ is monotone nondecreasing. Define $F(x)=\int^x_a f(t)\,dt$, $a\leq x\leq b$.

Let $\nu_G$ be the unique Borel measure on $(a,b]$ induced by $g$ (notice that $\nu_g$ is a nonnegative measure and that $\nu_g((x,y])=g(y)-g(x)$ for all $a\leq x\leq y\leq b$). Notice that $F$ is absolutely continuous. Let $\mu_F$ be the unique signed measure induced by $F$ (which is necessarily of finite variation). Then $\mu_F\ll m$ and $\mu_F=f\mathbb{1}_{[a,b]}\,dm$. An application of Lebesgue integration by parts yields \begin{align} \int^a_b f(t)g(t-)\,dt&=\int_{(a,b]}g(-t) \mu_F(dt)=F(b)g(b)-F(a)g(a)-\int_{(a,b]}F(t)\,\nu_g(dt)\tag{3}\label{three} \end{align} If $m=\min_{a\leq t\leq b}F(t)$ and $M=\max_{a\leq t\leq b}F(t)$, then $$m(g(b)-g(a))\leq \int_{(a,b]}F(t)\,\nu_g(dt)\leq M(g(b)-g(a))$$ If $g$ is not constant ($g(a)<g(b)$), there is $x_0\in[a,b]$ such that \begin{align} F(x_0)(g(b)-g(a))=\int_{(a,b]}F(t)\,\nu_g(dt)\tag{4}\label{four} \end{align} If $g$ is constant, any $x_0\in(a, b)$ will do. Substituting \eqref{four} on \eqref{three} yields \eqref{zero}: \begin{align} \int^b_a f(t)g(t-)\,dt&= g(a)(F(x_0)-F(a)) + g(b)(F(b)-F(x_0)) \end{align} \eqref{zero} follows immediately.

Remark: if in addition $f$ is not a constant function (a.s.) on $[a,b]$, then the point $x_0$ can be chosen in $(a, b)$.

• To obtained \eqref{one}, extend $g$ as $g(x)=0$ for $x<a$,and $g(x)=g(b)$ is $x>b$. Extend $f$ as $0$ out side $[a,b]$. By part (1), for any $\varepsilon>0$, there is $x_\varepsilon\in[a-\varepsilon,b]$ such that $$\int^b_af(t)g(t-)\,dt=\int^b_{a-\varepsilon}f(t)g(t-)\,dt=g(b)\int^b_{x_\varepsilon}f(t)\,dt$$ The conclusion follows by letting $\varepsilon\rightarrow0$

• \eqref{two} can be obtained from \eqref{two} by reversing the direction of the integrals, that is $\tilde{f}(t)=f(b+a-t)$ and $\tilde{g}(t)=g(b+a-t)$. Notice that $\tilde{g}$ is monotone nondecreasing whenever $g$ is monotone nonincreasing.