Prove that if $g(x)\leq g(y)$ then $F(x)\leq F(y),$ where $F(x)=\int^{g(x)}_{a}f(t)dt$

Question

Can you help me check if this is correct?

Prove that if $f$ is continuous and nonnegative on $[a,b]$ and $g:[a,b]\to [a,b]$is differentiable and nondecreasing on $[a,b],$ $$F(x)=\int^{g(x)}_{a}f(t)dt$$ is nondecreasing on $[a,b].$

PROOF 1

Since $g:[a,b]\to [a,b]$is differentiable and nondecreasing on $[a,b],$ then $g'(x)\geq 0,\;\forall\;x\in [a,b]$ \begin{align} F'(x)&=\dfrac{d}{dx}\int^{g(x)}_{a}f(t)dt \\&=f\left(g(x)\right)g'(x) \geq 0,\;\forall\;x\in [a,b]\end{align} Hence, $$F(x)=\int^{g(x)}_{a}f(t)dt$$ is nondecreasing on $[a,b].$

PROOF 2

Let $x,y\in[a,b]$ be fixed such that $x\leq y$. Since $g$ is nondecreasing, then $g(x)\leq g(y).$ Thus, $[a,g(x)]\subseteq [a,g(y)]$ and $$F(x)=\int^{g(x)}_{a}f(t)dt\leq \int^{g(y)}_{a}f(t)dt =F(y)$$ Hence, $$F(x)=\int^{g(x)}_{a}f(t)dt$$ is nondecreasing on $[a,b].$

Answer 1

This is correct; A little more general:

Claim. Let $[a,b] \subset\mathbb{R}$ be a interval. Let $f: [a,b]\rightarrow[0,\infty[$ be a Riemann-integrable function such that $x\le y \implies f(x) \le f(y)\ (\text{for all } x,y\in[a,b])$. Let $g: [a,b]\rightarrow[a,b]$ be an arbitrary function that satisfies $x\le y \implies g(x) \le g(y)\ (\text{for all } x,y\in[a,b]).\\\text{Then } \displaystyle F(x) := \int_a^{g(x)} f(t)\ \mathrm{d}t \text{ (for } x\in[a,b]\text{) is "nondecreasing".}$

Proof. Let $x,y\in[a,b]$ with $x\le y$. Then (by properties of the Riemann-Integral), \begin{equation*} F(y) := \int_a^{g(y)} f(t)\ \mathrm{d}t = \int_a^{g(x)} f(t)\ \mathrm{d}t + \int_{g(x)}^{g(y)} f(t)\ \mathrm{d}t \underbrace\geq_{\text{see (*)}} \int_a^{g(x)} f(t)\ \mathrm{d}t = F(x). \quad\square \end{equation*}

ADDENDUM \begin{align}\tag{*} & \text{Note that } \int_{g(x)}^{g(y)} f(t)\ \mathrm{d}t \overbrace\geq^{\text{property of Riemann-Integral}} (g(y)-g(x)) \cdot\inf_{x\in[a,b]} f(x) \geq 0,\\ & \text{since f }\geq0\text{ on }[a,b]\text{ and }g(y)\geq g(x)\text{ i.e. } g(y)-g(x)\geq0. \end{align}