The composition of a convex function with an integrable function is integrable If $f$ is (Riemann) integrable on $[0,1]$ and $\varphi$ is convex on $\mathbb{R}$, are there any simple or elegant proofs that $\varphi\circ f$ is also integrable on $[0,1]$?
This paper by J. Lu answers the question (and other similar questions) in a bit more generality--basically whenever $\varphi$ is merely continuous rather than convex--but I'm wondering if there's a simpler method here.
 A: This is not much of answer, but I just want to state that in my opinion the simplest proof of the result is the one the OP has already alluded to:

*

*Every convex function $\varphi$ is continuous.

*If $f: [a,b] \rightarrow \mathbb{R}$ is Riemann integrable and $\varphi: \mathbb{R} \rightarrow \mathbb{R}$ is continuous, then $\varphi \circ f: [a,b] \rightarrow \mathbb{R}$ is Riemann integrable.

The proof of 1) can be found in most "honors calculus" texts (e.g. Spivak).  If anyone wants a link to a proof, please let me know.  The proof of 2) given in the (nicely written, very elementary, expository) article by Jitan Lu cited by the OP works from the "Riemann notion of Riemann integrability", i.e., convergence of the Riemann sums uniformly in the mesh of the partition.  A lot of discussions of the Riemann integral (e.g. Spivak, Rudin) introduce the "Darboux notion of Riemann integrability" -- i.e., the one with the upper integral and the lower integral -- and with this notion the proof is shorter: closer to half a page than a full page.  See for instance the proof given on page 7 of these notes (Wayback Machine), which is taken almost verbatim from Russell Gordon's analysis text.
I suppose it would be of some interest if there were an even shorter proof that took advantage of the convexity, but in practice if you are at all interested in theorems like this you are going to want to know that $\varphi \circ f$ is Riemann integrable even for not necessarily convex continuous $\varphi$.
A: Jensen's theorem says
$$
\varphi\left(\int_If(x)\;\mathrm{d}x\right)\le\int_I\varphi(f(x))\;\mathrm{d}x
$$
where the measure of $I=1$. However, what you are asking is in the opposite direction and is not true. Consider the functions
$$
\begin{array}{}f(x)=x^{-\frac{1}{2}}&\text{and}&\varphi(x)=x^2\end{array}
$$
$f$ is integrable on $[0,1]$ and $\varphi$ is convex on $\mathbb{R}$, but $\varphi\circ f(x)=x^{-1}$ is not integrable on $[0,1]$.
On the other hand, if $\varphi$ were concave and non-negative, then Jensen's inequality insures that $\varphi\circ f$ is integrable.
Addendumb: As Pete Clark points out, and as has been shown here, the question must pertain only to proper Riemann integrals not converging improper integrals.
