(Wilson's integral Calculus 12.3.2-answered) Suppose I have that $f: [0,1]\to K$ is integrable, and I want to show that another function $g(x) = f(\sqrt{x})$ is also integrable in the same domain. How would I go about proving this?
I know that this is more suitable for using the definition that $A$ bounded function $f$ on $[a, b]$ is integrable if and only if for each $\epsilon>0$ there exists a partition $P$ of $[a, b]$ such that $$ U(f, P)-L(f, P)<\epsilon $$ . I tried to bound $ U(f(\sqrt{x}), P)-L(f(\sqrt{x}), P) <U(f, P)-L(f, P)<\epsilon$ but I am getting nowhere. By definition, $M(f, S)=\sup \{f(x): x \in S\} \quad \text { and } \quad m(f, S)=\inf \{f(x): x \in S\}$ by the definition I found in the book. How can I go about solving it? Any inputs are appriacted.
Take $\varepsilon>0$. You want to prove that there is a partition $P$ of $[0,1]$ such that$$U(g,P)-L(g,P)<\varepsilon.\tag1$$Since $f$ is integrable, there is a partition $P^\star=\{a_0(=0),a_1,\ldots,a_n(=1)\}$ such that $U(f,P^\star)-L(f,P^\star)<\varepsilon$. Let $P=\left\{a_0^{\,2},a_1^{\,2},\ldots,a_n^{\,2}\right\}$, and then you have $(1)$.