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Does Riemann integrability on closed interval implies uniform boundedness?

My thought process points to yes, because if f is Riemann integrable then it is bounded pointwise on [a,b]. I could be wrong, but I very vaguely remember from more elementary analysis that this implies uniform boundedness on that interval.

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  • $\begingroup$ Thanks, I was leaning towards that but was a little shaky. $\endgroup$ – Benjamin Lu Feb 25 '13 at 0:24
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    $\begingroup$ What do you mean by "uniform boundedness" of a single function? Just that it is bounded on $[a,b]$? The answer then is "yes". $\endgroup$ – David Mitra Feb 25 '13 at 0:25
  • $\begingroup$ ^Yeah that was what I meant. $\endgroup$ – Benjamin Lu Feb 25 '13 at 0:27
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    $\begingroup$ @Isaac: that can be posted as an answer. $\endgroup$ – Willie Wong Feb 28 '13 at 1:12
  • $\begingroup$ What do you mean by "pointwise bounded" and "uniformly bounded" for a single function? That is very confusing terminology. $\endgroup$ – Pete L. Clark Jul 26 '13 at 20:10
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The definition of Riemann integral involves taking arbitrary sample points in subintervals, or supremum or infimum over said subintervals. Either process fails to arrive at a finite value if the function is unbounded. Thus, a Riemann integrable function must be bounded.

That said, the improper Riemann integral is defined in greater generality, and it can be finite for some unbounded functions such as $x^{-1/2}$ on $[0,1]$.

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A continuous, pointwise bounded function is uniformly bounded on any compact set $X$, as can be seen by considering the cover $X_n=\{x \in X:|f(x)|<n\}$.

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    $\begingroup$ This only works if we assume $f$ is continuous, which need not be the case. $\endgroup$ – Alex Becker Jul 26 '13 at 19:55

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