Recent content by canis89

Yes, it is. In fact, you can find the integral of f from 0 to 1 just by integrating both sides of the equation defining f.

Spitz, as micromass have told you, you've already had the right answer from the beginning. Sorry for being so difficult. It maybe partially due to me having a flu (damn the weather!). Anyway, I was trying to help you prove that it is the right answer by yourself. Anyway, cheers. Oh, and...

Sorry, I don't think I can make sense of what you've just said. Anyway, to show that f is measurable with respect to {emptyset, (-infinity,0), [0,infinity), R}, you have to show that: For every lambda, {x:f(x)>lambda} is a member of {emptyset, (-infinity,0), [0,infinity), R}, i.e., you have...

Ok, then. I believe you're good on track. Just work on f measurability now.

Hmm, let me get this straight. If M is a sigma-algebra of R (real space), then, M must be a subset of the power set of R. That is, M is of the form {emptyset, R, and some other subsets of R satisfying the former two conditions} (M is a set of sets, while R is a set of numbers). Is this what you...

Do you mean by R as the element of the sigma-algebra? I'll assume so. Yes, you're correct for those two conditions. You're missing one though. A sigma-algebra M of a space X must include the emptyset and X itself. So, have you shown whether...

Spitz, the definition of sigma-algebra is what you need first. If you can't find it in your notes, you can search wikipedia or wolfram for this. If you want to make sure, you can present the definition here so we can check it. So, I'll just comment about measurability here. If you're working...

Hint: What must be the values of a,b\in\mathbb{R} so that a^2+b^2=0?

Spitz, to answer your question, you have to first ask whether \left\{\emptyset,(-\infty,0),[0,\infty),\mathbb{R}\right\} is a sigma-algebra. If it is a sigma-algebra, check whether f is measurable with respect to this sigma-algebra. If yes, can you reduce the set such that the reduction is still...

But then, T will violate all three conditions. Notice that im(T)={(0,y,z)}; im(T^2)={(0,0,z)} ker(T)={(0,0,z)}; ker(T^2)={(0,y,z)} Then, i) im(T) (+) ker(T)={(0,y,z)} is pure subset of R^3 ii) im(T^2) is pure subset of im(T) iii) ker(T) is pure subset of ker(T^2)

Hi amanda, about the first part, you only show that f is continuous on p. If you want to show that f is continuous on X, then you have to show that for any x_0\in X, for every \epsilon>0, there exists \delta>0 such that d(x,x_0)<\delta implies |f(x)-f(x_0)|<\epsilon. Or maybe you only want to...

Hi, this kind problem has been examined very thoroughly by Mike Spivey at http://math.stackexchange.com/questions/73758/probability-of-n-successes-in-a-row-at-the-k-th-bernoulli-trial-geometric The discussion covers the derivation of the distribution's pmf and PGF.

Oh, I just want to make sure that when you reduce M, the result must contain only the finite subfamily of A' (it must not contain F complement). But it is straightforward since A=E\cap F cannot be a subset of F complement.

I'll try to help for 1a) first. By your assumption, f maps X to Y. So, an inverse image of f must be a subset of X, not Y(or maybe you did a typo there?). Hint 1. Try to relate the statement f is continuous on Ui with the statement (ii): Hint 2. Since X is composed of Ui's, the...

Yes, you're right. That's the idea. Forget about the union of A' and the complement of E. It's redundant. You only have to union A' with F complement. Now, can you write your argument more specific?