Is the function $f(x)=\sin(1/x)$ Lebesgue integrable on $(0,1]$?
I know that, as $f$ is continuous on the set, it is a measurable function. However, I'm stumped on how to go on. A nudge in the right direction would be greatly appreciated.
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$\begingroup$ It is even Riemann integrable. $\endgroup$– Marc van LeeuwenCommented Feb 10, 2015 at 11:17
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$\begingroup$ Does Riemann imply Lebesgue on finite intervals then? $\endgroup$– OriCommented Feb 10, 2015 at 11:19
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1$\begingroup$ Yes $\endgroup$– Marc van LeeuwenCommented Feb 10, 2015 at 11:27
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1 Answer
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Hint: $|f(x)|\le1$ for all $x\in(0,1]$, so $f$ is uniformly bounded.
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$\begingroup$ Thanks, I can't believe I missed something so obvious. $\endgroup$– OriCommented Feb 10, 2015 at 11:15
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1$\begingroup$ What does it mean for a single function to be uniformly bounded? $\endgroup$ Commented Feb 10, 2015 at 11:20
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$\begingroup$ Yeah alright, "bounded" was probably enough. $\endgroup$– JasonCommented Feb 10, 2015 at 12:02