Recent content by QIsReluctant

I think I've really got it this time, thanks to some help from someone who had the same question. Suppose ∫01 g(x)dx < ∞. Then g is Lebesgue integrable since it is positive; thus the Dominated Convergence Theorem implies that ∫[0, 1/n] g(x)dx → 0. But this contradicts the hypothesis that says...

Are you willing to work harder than you had been working to get the 2.8? If not, then forget it; you won't make your GPA better and could possibly make it worse. If yes, then check the math to make sure you can get 3.0 or better with As, commit to being more diligent, follow through, and, if...

I gave it a little bit of thought before going to bed last night, and the answer is now obvious Since ∫[0,1/n] fn(x)dx > (n-1)/n, we have that sup[0,1/n] fn(x) ≥ n - 1. But since g(x) ≥ fn(x) for all x and n, we have that ∫[1/n, 1/n+1] g(x)dx ≥ (n - 1)[ 1/n - 1/(n+1) ]. We represent ∫[0,1]...

No -- I meant that, at each individual x, we choose the supremum of the set of values at x. For example, if F = {the functions x → ax: a ∈ (0, 1)}, then g(0.5) = 0.5 and g(-0.5) = 0.

Homework Statement Let X be a compact metric space, F a family of uniformly bounded, equicontinuous real-valued functions on X. Is the function g(x) := supf ∈ F f(x) necessarily continuous? The Attempt at a Solution The hypotheses about the function family seem to point to some use of...

Ach, I'm really not getting anywhere. Would you mind being more detailed? The biggest problem for me is that we have no (apparent) guarantee that fn(x) will ever hold some minimum value anywhere. So g(ε) might be zero for all we know. I can find plenty of lower bounds for sup f on [0, 1/n]...

Not necessarily so -- that would only be true if the sup function were constant. Or have I just misunderstood?

Thank you!

I proved that the integrals exist and are the same for positive, measurable functions. So if g is a general measurable function, then we have g = g+ - g-. Take the integral over ℝ. Each of the two integrands (of ∫ g+dμ and ∫g-dμ ) represents a positive, measurable function, so we can replace...

Okay, I think I am starting to make progress here. A few questions: (1) We can assume that the a_i are positive because, if not, we're assuming the simple function is positive and so could just replace the simple function with another one with positive coefficients, right? (2) Does "exist" in...

Aha! That's the reason. Glad I could help.

Homework Statement For n = 1, 2, ..., let fn be a Lebesgue integrable function [0,1] → [0, +∞) such that (1) ∫01 fn dx = 1 and (2) ∫1/n1 fn dx < 1/n Let g(x) = supn ∈ ℕfn(x). Prove ∫01 g(x)dx = +∞ The Attempt at a Solution Coffee, banging my head against a wall, etc. There's not enough...

But, why would they not simplify this further so that it is i/4? Because that would not be a valid simplification. 1/i = -i, and so we would need -i/4 and not i/4. I'd be interested in seeing this Wolfram Alpha output you speak of ...

Looks like a simple algebra mistake. When you simplify for the second time you eliminate the negative sign, effectively dividing by -1. That's why your answer differs by a factor of -1.

Homework Statement Let g: ℝ → ℝ be Lebesgue measurable, let μ(A) be a measure of Lebesgue measurable sets defined by μ(A) = ∫A f dx, where f: ℝ → ℝ is non-negative, with finite integral on compact intervals. Prove that ∫ℝ g(x)dμ(x) exists if and only if ∫ℝ g(x)f(x) dx exists, in which case...