Recent content by WHOAguitarninja

Figured it out. If we define Fn to be f if f<fn and fn if fn<f, then we have that |Fn|<|f|+|fn|, which we is integrable and the limit of it's integral is 2 times the limit of the integral of |f|. Since Fn converges a.e. to f, we then have that the integral of f is equal to the limit of the...

After poking at it a bit more, if I could establish the same inequality that I'm arriving at on my last step for the portions of f and fn where f>=fn and where f <=fn then I could get it from there. (that is, split |f|-|fn| into positive and negative portions.)

Homework Statement This comes courtesy of Royden, problem 4.14. a.Show that under the hypothesis of theorem 17 we have \int |fn-f| \rightarrow 0 b.Let <fn> be a sequence of integrable functions such that fn \rightarrow f a.e. with f integrable. Then \int |fn-f| \rightarrow 0 if and only...

This is a statement my professor made in class some time ago (as a means to show that C contains a Hamel basis) that seemed fairly innocent, but it's bothered me for awhile. I did some searching online, and it seems that C+C=[0,2]. There it was again stated that this is fairly easy to show...

Be careful, as morphism is saying, what you are talking about is the outer measure, NOT the measure. A non measurable set still has outer measure. In any case, have you looked at the Vitali non measurable set? You may be able to construct an example if you consider that.

Try this...pull a 1/h to the outside, and consider how you might be able to rewrite the fraction as two more useful fractions added together (or subtracted).

So let me see if I understand the restriction correctly, as I don't think the book spells this out very well. We can only state that it's 1-1 if V is some, perhaps restricted, neighborhood of the image of U. That much I see. However we also must restrict U to the neighborhood in which...

Aye...title should say in R^2, sorry about that. I'm hitting somewhat of a wall in my understanding of a theorem (or rather a special case of a theorem). The theorem as stated in the book is as follows. The Inverse FUnction Theorem in the Plane Let O be an open subset of the plane R^2 and...

This is what I was thinking as well. I suspect it.anoopk's solution was probably the original intention. Seems a bit less random.

Ah, good call, even simpler than I thought, thanks!

Prove that if the functions g:[a,b] --> R and h:[a,b] --> R are continuous, with h(x)\geq0 for all x in [a,b] then there is a point c in (a,b) such that \int h(x)g(x)dx = g(c) \int h(x)dx when the integrals go from a to b --- My immediate thought was Integral MVT, so I said \int...

You know...it's funny...I had considered multiplying by sqrt(n+1)+sqrt)n), but decided against it since it limits to infinity and I figured it wouldn't tell me anything...didn't even consider just multiplying what you suggested. Thanks a million.

Homework Statement Discuss the convergence of the following sequences at infinity. a)\sqrt{n+1}-\sqrt{n} b)\sqrt{n}(\sqrt{n+1}-\sqrt{n}) c)n(\sqrt{n+1}-\sqrt{n})I've already solved a, and if I can solve b then I have c automatically. This is for an undergraduate advanced calc course so all...

Homework Statement This is the result of a problem from my Quantum class, but I figure it would be best to ask in here as my question is purely a question of how to solve a certain differential equation. the equation is of the form 0=Y''-i*a*Y' + b*Y where Y is a function of t So the...

Thanks. That makes perfect sense. I wasn't thinking more along the lines of showing a-d were not groups than were not subgroups. I suppose it then makes sense that I could never explain that they weren't, because they are! Thanks a lot, that really helped.