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I have shown that $f(x)=\sin(1/x)$ is Riemann integrable on $(0,1]$, but I am wondering if $f(x)=(\sin (1/x))^4$ is Riemann integrable on $(0,1]$?

It isn't hard to show that $\sin(1/x)$ is Riemann integrable on $(0,1]$.

Proof. Given $\epsilon > 0$, consider the partition $[0, \epsilon]$ and $[\epsilon, 1]$. Since $f$ is bounded and continuous on $[\epsilon, 1]$ it is integrable, so there is a partition $P_1$ such that $U - L < \epsilon$. On $[0, \epsilon]$, the most $U - L$ could be (for any partition, $P_\epsilon$) is $2 \epsilon$ (since $\max_{x \in \mathbb{R}} \sin x = 1$, $\min_{x \in \mathbb{R}} \sin x = - 1$, and length of the interval is $\epsilon$). Now consider the partition of $[0, 1]$ given by $P_1$ union $P_e$ and look at $U - L$. We have $|U - L|$ on $[0, 1]$ is less than or equal to the sum of $U - L$ for the two partitions $P_1$ and $P_e$, which is less than $\epsilon + 2 \epsilon = 3 \epsilon$. So $U - L < 3\epsilon$, which means it can be made arbitrarily small. Thus $f$ is integrable on $[0, 1]$.

Is there any way I can adapt this proof to address $f(x)=(\sin (1/x))^4$?

Clyde Kertzer
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    Riemann integration is defined on closed bounded intervals. improper Riemann integrals are defined on more general intervals by passing to to the limit on certain Rieman integrals. In your case, you may defined $f$ as whatever you want at $x=0$ to talk "properly" about a Riemann integral. In any event, $f$ is indeed Riemann integrable on $[0,1]$ (regardless of what (finite) value is assigned to $f$ at $x=0$. Multiplication of functions that are Riemann integrable over an interval $[a,b]$ are also Riemann integrable, – Mittens Apr 20 '22 at 04:35
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    @Clyde Kertzer Your proof makes use of the property that $f(x)$ is bounded and continuous on $(0,1].$ Hence it captures $\sin(1/x),$ as well as $(\sin(1/x))^4.$ You do not need any additional theorems. – Ryszard Szwarc Apr 20 '22 at 06:01

2 Answers2

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Any function that has measure 0, of discontimuities is Riemamn integradable . The above mentioned function is such.

dmtri
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    That is correct, but this deep result by Lebesgue is very likely out of the OPs reach. The intergbility of $f$ can be established by more pedestrian methods accessible to Calculus students. The rest is just the fact that multiplication of Riemann integrable functions over an interval $[a,b]$ are also Riemann integrable. – Mittens Apr 20 '22 at 04:41
  • You are 100 per cent right. I do not expect an accepted answer here. If you think the answer and also these comments are useless, I will delete it .@OliverDíaz – dmtri Apr 20 '22 at 04:53
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    Not at all, It was a friendly criticism. I was just making the point that your kung-fu is probably to advanced to our young grass hopper. – Mittens Apr 20 '22 at 04:59
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    Yes, unfortunately if I used measure justification, I would not receive any credit. But, I appreciate the response, so I will give you an upvote nonetheless. – Clyde Kertzer Apr 20 '22 at 05:08
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The product of two Riemann integrable functions is Riemann integrable function. The product of two Riemann integrable functions is integrable. $f(x)^4$ can be written as a product of Riemann integrable functions.

ganesh
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