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I need some help here, folks...

Problem: If $f$ and $g$ are integrable on $[a,b]$ and $g(x) \leq f(x)$ for all $x \in [a,b]$, then $\int_a^b g \leq \int_a^b f$.

Any help would be great! Thanks!

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Do you know a theorem that says something along the lines of, if $h$ integrable and $h\geq 0$, then $\int h \geq 0$? – Matt Nov 28 '12 at 18:20
Is this a Riemann integral question or Lebesgue? – Gautam Shenoy Nov 28 '12 at 18:21
If not, use the definition of (I'm assuming) Riemann integration, showing that the upper bound on any interval within a partition of $f$ is greater than $g$... – Emily Nov 28 '12 at 18:21

I'll assume it is a riemann integral. Let $f \geq 0$. Observe that the Riemann upper sum and lower sum for this is $\geq 0$. (From Rudin), we define $$\underline{\int_a^b} f dx = \sup_{\mathbb{P}} \sum m_i \Delta x_i$$ $$\overline{\int_a^b} f dx = \inf_{\mathbb{P}} \sum M_i \Delta x_i$$ where $m_i$ and $M_i$ are the infimum and supremum respectively of the intervals decided by the partitions. Please read the Riemann integrals chapter from Rudin to understand what I am saying. My point is that both the integrals mentioned above are non -ve and so the riemann integral which is equal to both as it exists is non negative. Now substitute f with f-g, use linearity and you get your result.

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Keeping technical details out of this because it is a homework problem, but please ask where you would need clarification.

Assuming ordinary Riemann Integration, decompose the $\int f$ into a sum of rectangles over a partition. Every $f$-rectangle must be at least as high as the corresponding $g$-rectangle, hence the same limit for $g$ would converge to a smaller value.

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The (Riemann) integral of $f$ cannot be decomposed into a sum of rectangles over a partition. – Did Nov 28 '12 at 18:34
Thank for everyone's help! I was on the same page as everyone, but I just needed some validation. – user42864 Nov 28 '12 at 18:41

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