Composition of Riemann integrable function and C1 function is Riemann integrable

Question

Let $f$ be a Riemann integrable function on $[a,b]$ and $g: [\alpha, \beta] \to [a, b]$ be a continuously differentiable function. Then how can I prove that $f \circ g$ is Riemann integrable? I think this is an easy task, but I can't get it. Maybe uniform continuity of $f$(+uniform continuity of $f'$) seems to be crucial, but I can't proceed any more.

There are no more assumptions about $f$ other than stated. According to the textbook containing this(Krantz, Real Analysis and Foundations, 4th ed, sec 7.3. It is an elementary real analysis textbook), this is an 'easier result' than proving that 'composition $\phi \circ f$ of continuous function $\phi$ on compact interval and Riemann integral function $f$ on $[a,b]$ is Riemann integrable'. I would appreciate answers not using measure theory if possible.

Answer 1

This is false. Let $\alpha=a=0, \beta = b = 1.$ Define $f=\chi_{(0,1]}.$ Then $f$ is Riemann integrable on $[0,1].$

To define $g,$ we use the following fact: Given a closed subset $E\subset \mathbb R,$ there is a $C^\infty$ function $g:\mathbb R\to [0,1]$ such that $g=0$ on $E$ and $g>0$ everywhere else.

With that in mind, choose a nowhere dense compact set $K\subset [0,1],$ with $m(K)>0.$ Use $E=K$ in the last paragraph to obtain $g\in C^\infty$ with the properties above.

Claim: $f\circ g$ is discontinuous at each $x\in K.$

Proof: Let $x\in K.$ Because $K$ is nowhere dense, there exists a sequence $x_n \in [0,1]\setminus K$ with $x_n\to x.$ We then have $g(x_n)>0$ for all $n.$ Hence $f\circ g(x_n) = 1$ for all $n.$ But $x\in K$ implies $g(x)=0,$ which gives $f\circ g(x)=0.$ This gives the claim.

Now if $f\circ g$ were Riemann integrable on $[0,1],$ then its set of discontinuities would have measure $0.$ But the claim implies $f\circ g$ is discontinuous on a set of positive measure, namely $K.$ This is a contradiction, proving $f\circ g$ is not Riemann integrable

Answer 2

This is true if in addition to $g \in C^1([\alpha,\beta]$), the derivative $g'$ is non-zero. We then have an inverse function $g^{-1}$ which is continuously differentiable, mapping sets of measure zero into sets of measure zero.

Since $f$ is Riemann integrable it is bounded and the set of discontinuity points $D_f$ has measure zero. Thus, $f \circ g$ is Riemann integrable since it is bounded with discontinuity points in a set of measure zero, $D_{f \circ g} = g^{-1}(D_f)$.