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If the Riemann sums of a function $f$, defined on an arbitrarily tagged partition $P$, are positive, then will the integral also be positive for any arbitrary tagged partition whose norm is bounded?

Michael Albanese
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Raghav
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    If the function is integrable, yes. Your hypothesis implies that $f$ is positive everywhere. If $f$ is integrable, the integral is then necessarily positive. See this for example. (Or argue using the fact that the set of points where a bounded, integrable function is discontinuous has measure zero; $f$ will then have at least one point of continuity. The result follows easily from this). – David Mitra Jun 03 '13 at 15:36
  • In the above, I assumed you meant to ask "will the integral also be positive?". – David Mitra Jun 03 '13 at 15:39
  • Can this not be argued using more elementary methods,perhaps from the definition itself ? More precisely, If $S(f;\dot(P))$ is the Riemann sum on any arbitrarily tagged partition $\dot(P)$ of $[a,b]$ ,such that $\forall\epsilon>0,\exists \delta_{\epsilon}>0$,such that if $|\dot(P)|<\delta_{\epsilon}$,then $|S(f;\dot(P))-\int^b_a f|<\epsilon$. Using this if we fix $\epsilon:=\frac{S}{2}$,where $S$ is $S(f;\dot(P))$,then for a certain $\delta_{\epsilon}$,such that if $|\dot(P)|<\delta_{\epsilon_{0}}$,then$ 0<\frac{S(f;\dot(P))}{2}<\int^b_a f$. Can such an argument suffice ? – Raghav Jun 03 '13 at 16:47
  • Admittedly,this argument is inspired from a similar question in continuity.Hence the doubt regarding the validity of its usage. – Raghav Jun 03 '13 at 16:52
  • I'm not following your argument. It seems off... But, here is another relevant post. This post and the linked posts theiren (in particular, the first answer here contains arguments more direct than those I mentioned in my first comment. – David Mitra Jun 03 '13 at 17:10

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