Thread:

1. Integrable Functions

Let f be a function on the interval [a, b] such that for every epsilon > 0 there exist integrable functions g, h on [a, b] with h(x) <= f(x) <= g(x) and ((integral from a to b g(x)dx) - (integral from a to b h(x)dx)) < epsilon. Show that f is integrable on [a, b].

Any ideas on where to start and/or how I would go about showing this? Thanks.

2. Riemann integrable?

Assuming that, take some partition of [a, b] and look at the Riemann sums for g, f, and h. You should be able to show that [itex]R(g)\le R(f)\le r(h)[/tex] (R(f) is the Riemann sum of f for the particular partition).

"((integral from a to b g(x)dx) - (integral from a to b h(x)dx)) < epsilon" is a bit peculiar since the two integrals are specific numbers. That really just says the integrals of g and h are the same. That is that the lim, as the number of intervals in the partition goes to infinity, of the Riemann sums, R(g) and R(h), are the same so the limit of R(f) exists and is that joint value.

3. Originally Posted by kjwill1776

Let f be a function on the interval [a, b] such that for every epsilon > 0 there exist integrable functions g, h on [a, b] with h(x) <= f(x) <= g(x) and ((integral from a to b g(x)dx) - (integral from a to b h(x)dx)) < epsilon. Show that f is integrable on [a, b].

Any ideas on where to start and/or how I would go about showing this? Thanks.
$\int h \leq \int_{-} f \leq \int g$ and $\int h \leq \int^{-} f \leq \int g.$ thus $0 \leq \int^{-}f - \int_{-}f \leq \int g - \int h < \epsilon.$ since $\epsilon > 0$ is arbitrary, we must have $\int^{-}f - \int_{-}f=0.$
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