If $g$ is Riemann-integrable in a closed interval and $f$ is a increasing function in a closed interval, is $g\circ f$ Riemann-integrable?
To clarify: the problem stated that the composition is well defined. I think that the statement of the question is true but Im having trouble to write/concrete a proof.
I know that a monotone function defined in a closed interval is Riemann-integrable by a previous result. And I know that the reverse statement ($g$ being monotone and $f$ integrable) is not true in general.
After thinking some moment my idea for the proof is to show that $g\circ f$ have, at most, countable discontinuities. To do this I was thinking how to show some kind of order-correspondence between the points of the domain of a monotone function and it image.
Can you help me? I get stuck and it is very possible that Im wrong in my assumption about that the statement is true. Some hint will be appreciated.
EDIT: I get a new idea, design a partition of $f$ that take any possible discontinuity into a closed interval of known length, and after see what happen in the composition with this closed intervals with discontinuities, and by the other hand see what happen with the parts of continuous mapping.
After reading this answer I get the idea to provide a proof for the validity of the statement.
First I will characterize the different kind of images of closed intervals that a monotonic function can create. For an increasing function we have that $x<y$ implies that $f(x)\le f(y)$, and that at most a monotone function can have a countable number of discontinuities in any closed interval.
No discontinuities on the image: if the image of $f([x_1,x_2])$ is continuous then it can be at most of three types:
A constant function $f([x_1,x_2])=\{a\}$. Then $(g\circ f)([x_1,x_2])=g(\{a\})=\{c\}$, then the function $(g\circ f)$ is constant in $[x_1,x_2]$ so is Riemann-integrable
A strictly increasing function i.e. $f([x_1,x_2])=[a,b]$. Then $(g\circ f)([x_1,x_2])=g([a,b])=[c,d]$. If $g$ is Riemann integrable in $[a,b]$ then exists a sequence of partitions $(P_n)$ such that
$$\lim_{n\to\infty}U(g,P_n)-L(g,P_n)=0$$
Then because $f$ is bijective in $[a,b]$ then for every $P_n$ exists a partition $P'_n$ in $[x_1,x_2]$ such that
$$\lim_{n\to\infty}U(g\circ f,P'_n)-L(g\circ f,P'_n)=0$$
so $g\circ f$ is Riemann-integrable in $[x_1,x_2]$
- A mix of both previous cases. Then the interval $[x_1,x_2]$ can be partitioned on types of subintervals discussed previously (constant and strictly monotonic images), so $g\circ f$ is Riemann-integrable here too.
Discontinuity in the image: a jump discontinuity in $[x_1,x_2]$ is mapped into two types of the previously discussed intervals with the difference that can be the case that in the image exists an interval with an open boundary of the kind $[a,b)$ or $(a,b]$.
But cause $f$ and $g$ are Riemann integrable in closed intervals this mean that they are bounded, and if $g$ is integrable in any subinterval $[a-\varepsilon,b]$ then is integrable in $(a,b]$ (this is known by a previous proof).
My question: it is this proof correct? It lacks something essential? Can you help me to write it better? Thank you in advance.
EDIT 2: as the user @ParamanandSingh pointed in the comments it is possible that the discontinuities cannot be isolated, one by one, inside some interval. So I will search further for a correct proof of the statement.