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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.

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Should clarify this is about Riemann, not Lebesgue integral. – scineram Sep 30 '11 at 8:19
Now I cannot access the paper, but is it not trivial after looking at the integrability criterion of almost everywhere continuity? – scineram Sep 30 '11 at 8:26
Yes, it is trivial if you use that criterion, but that seems a bit heavy duty for something like this. I feel like there should be a more direct and enlightening approach. Also, this is just the Riemann integral. – Zach Conn Sep 30 '11 at 13:59
Hi Zach, It is not clear for me why the scineram comment settle the question. Because of the set of discontinuity of $\varphi\circ f$ is clearly a subset of the set of discontinuity of $f$, but why this is a Lebesgue measurable set ? – Leandro Sep 30 '11 at 15:37
Does it help if $\varphi$ is Lipschitz on every bounded set? – Niels Diepeveen Oct 1 '11 at 0:25
up vote 2 down vote accepted

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:

1) Every convex function $\varphi$ is continuous.
2) 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, 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 $\varphi$.

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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.

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$@$robjohn: Your $f(x)$ is not bounded on $[0,1]$, so it cannot be Riemann integrable. It has a convergent improper integral, but that is not the same thing and -- evidently -- the result in question is meant to apply to such functions. – Pete L. Clark Oct 1 '11 at 0:01
"not meant", I should have said. – Pete L. Clark Oct 1 '11 at 0:10
@Pete: ah, okay. So $\varphi\circ f$ is bounded on $[0,1]$. Not much to show. $\varphi$ is either monotonic or monotonic decreasing then monotonic increasing. Easy to show that $\varphi\circ f$ satisfies the other conditions for Riemann integrability. – robjohn Oct 1 '11 at 0:11
@Pete: my answer has been so annotated. – robjohn Oct 1 '11 at 0:22

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