I'd love your help with the following question: I need to prove or refute the claim that for a Riemann integrable function $f$ in $[0,1]$ also $\sin(f)$ is integrable on $[0,1]$.

My translation for this claim: If $\int_{0}^{1} f(x) dx < \infty$, so does $\int_{0}^{1} \sin(f(x))dx < \infty$, Am I right?

I tried to think of an elementary function that will fit the conditions, one that will blow up in $0$ or $1$ or both, but I didn't find any. Can I just use the fact that $\int_{0}^{1} \sin(f(x))dx \leq \int_{0}^{1} 1dx < \infty$ and that's it or Am I missing something?


  • $\begingroup$ but if we are with the Riemann eyes, that's not enough. $\endgroup$ – leo Feb 10 '12 at 18:10
  • $\begingroup$ So my translation is incorrect indeed, what am I miss? $\endgroup$ – Jozef Feb 10 '12 at 18:12
  • $\begingroup$ I remember from calculus a hard to prove exercise that says that if $f$ is continuous and $g$ is Riemann-Integrable on $[a,b]$ then $f\circ g$ is Riemann-integrable over $[a,b]$. $\endgroup$ – leo Feb 10 '12 at 18:13
  • $\begingroup$ Is not incorrect. Seems like you are trying to prove that the function $\sin(f)$ is Lebesgue integrable over $[0,1]$. Indeed, it is, but that don't implies Riemann-integrability of $\sin(f)$ on $[0,1]$ $\endgroup$ – leo Feb 10 '12 at 18:18

Since the sine function is $1$-Lipschitz continuous, the difference between the upper and the lower Darboux sums based on a given pointed subdivision for $\sin(f)$ is at most the difference between the upper and the lower Darboux sums based on the same pointed subdivision for $f$. Since $f$ is Riemann-integrable, the latter can be made as small as one wants. Hence the former can also be made as small as one wants, which is equivalent to the Riemann-integrability of $\sin(f)$.

  • $\begingroup$ I prefer this one $\endgroup$ – leo Feb 10 '12 at 18:43
  • $\begingroup$ +1: This is exactly the approach I took last week in my Spivak calculus course: see $\S 3.4$ of math.uga.edu/~pete/2400int.pdf for how it went. $\endgroup$ – Pete L. Clark Feb 11 '12 at 2:43
  • $\begingroup$ @DidierPiau I always preferred the Darboux sums Spivak introduces, though he never talks about Darboux. $\endgroup$ – Pedro Tamaroff Feb 11 '12 at 13:04

I do not think your translation is correct (unless you meant Lebesgue, and not Riemann integrable). The concept of Riemann integrable and Lebesgue integrable are not the same.

Riemann integrable: $f\colon[a,b]\to\mathbb{R}$, $-\infty<a<b<+\infty$, $f$ bounded and the upper integral equal to the lower integral.

Lebesgue integrable: $f\colon E\subset\mathbb{R}\to\mathbb{R}$, $E$ Lebesgue measurable set, $f$ Lebesgue measurable and $\int_E|f|<\infty$.

There are several ways of showing that if $f\colon[0,1]\to\mathbb{R}$ is Riemann integrable so is $\sin(f)$. The easiest way is to use the following fact: a bounded function $g\colon[a,b]\to\mathbb{R}$ is Riemann integrable if and only if the set of points where $g$ is discontinuous has measure $0$.

First of all it is cleat that $\sin(f)$ is bounded. Since $\sin$ is a continuous function, $$ \{x\in[0,1]:\sin(f)\text{ is discontinuous at }x\}\subset\{x\in[0,1]:f\text{ is discontinuous at }x\}. $$ Since $f$ is Riemann integrable, the set on the right hand side is of measure $0$, and so are all its subsets.

  • $\begingroup$ "has measure 0" means that there א0 points where the function is not continuous? $\endgroup$ – Jozef Feb 10 '12 at 18:27
  • $\begingroup$ The answers is fine. But if this is an exercise for one of the firsts analysis course, perhaps the Lebesgue theorem is not covered and then can't be used. $\endgroup$ – leo Feb 10 '12 at 18:28
  • $\begingroup$ It was covered, but never used. It's more Calculus part 3 I believe. $\endgroup$ – Jozef Feb 10 '12 at 18:34
  • $\begingroup$ @Jozef, a set $A$ have measure $0$ if for a given $\epsilon\gt 0$ there exist a countable collection of intervals $\{(a_k,b_k)\}$ such that $A\subseteq \bigcup (a_k,b_k)$ and $\sum (b_k-a_k)\lt \epsilon$. Sets of measure theory can be introduced with any knowledge of measure theory, however Lebesgue integrability condition requires measure theory. $\endgroup$ – leo Feb 10 '12 at 18:35
  • $\begingroup$ @Jozef, in "sets of measure theory" read "sets of measure $0$", and in "with any" read "without any" $\endgroup$ – leo Feb 10 '12 at 18:44

Assuming you mean Riemann integrable everywhere, recall:

$f$ is Riemann integrable over $[a,b]$ if and only if $f$ is bounded and for every $\epsilon>0$, there is a partition of $[a,b]$ such that $U_f(a,b) -L_f[a,b]<\epsilon$, where $U_f$ and $L_f$ are respectively the upper and lower Riemann sums of $f$ corresponding to the partition.

So, to show that $f$ is Riemann integrable on $[a,b]$ you have to demonstrate that it is bounded, and you have to show that the above condition concerning upper and lower sums holds. Towards this end, you can argue as in Didier Piau's answer. (alternatively, see Julián Aguirre's answer.)

Your argument is flawed, because you're essentially assuming that the expression $\int_a^b \sin(f(x))\,dx$ makes sense to begin with.

It is easy to show that $\sin(f(x))$ is bounded; but to show it's integrable, you can't immediately write "$\int_a^b \sin(f(x))\,dx<\infty$". Either $\sin(f(x))$ is integrable, in which case the definite integral is finite, or it's not, in which case the expression $\int_a^b \sin(f(x))\,dx$ has no valid meaning.


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