Dense subset

**derek walcott** · April 1st 2010, 07:06 AM

Let A be a dense of [a,b] and let f:[a,b]-->R be an integrable function such that f(x)>=0 for every element of x in A. Prove that the (integral of a to b) f >= 0.

Please help. I have ideas for this but need help solving

**Tinyboss** · April 1st 2010, 08:26 AM

I assume you mean f is Riemann-integrable--it isn't true for Lebesgue integration.

For any partition P, every interval will contain an element of A, and therefore the upper sum is non-negative for that partition. Since f is integrable, the upper sums converge to the integral as the mesh goes to zero, and so the integral is non-negative.

**Drexel28** · April 1st 2010, 08:47 AM

Originally Posted by derek walcott

Let A be a dense of [a,b] and let f:[a,b]-->R be an integrable function such that f(x)>=0 for every element of x in A. Prove that the (integral of a to b) f >= 0.

Please help. I have ideas for this but need help solving

Originally Posted by Tinyboss

I assume you mean f is Riemann-integrable--it isn't true for Lebesgue integration.

For any partition P, every interval will contain an element of A, and therefore the upper sum is non-negative for that partition. Since f is integrable, the upper sums converge to the integral as the mesh goes to zero, and so the integral is non-negative.

Another way to say this depending your def. of Riemann integrable. As Tinyboss said $\sup_{x\in I\subseteq[a,b]}f(x)\geqslant 0$ . Thus, $U(P,f)\sum_{j=1}^{n}\sup_{x\in[x_{j-1},x_j]}f(x)\Delta x_j\geqslant 0$ regardless of $P$ . Clearly then $0\geqslant U(P,f)$ for every partition $P$ in $\mathcal{P}[a,b]$ . Thus, $\int_a^bf=\inf_{P\in\mathcal{P}[a,b]}U(P,f)\geqslant 0$

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