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Real Analysis Q / Sigma-algebra

Suppose X is a measurable space, E belongs to the sigma algebra ( I believe to the sigma algebra in X) , let us consider X\E = Y. Show that all sets B which can be expressed as A\E, where A belongs to the sigma algebra in X, form a sigma-algebra in Y. Please justify every step and claim you make in the solution. I know this problem is straight forward from the def of sigma-algebra, but I want to double check my answer.

Subject:

Math

Topic:

Real Variables

Posting ID:

47937

OTA ID:

103300

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Explanation of the condition - not independent of the Jacobians of functions.

Real Analysis Jacobians(I) Explanation of the condition - not independent of the Jacobians of functions.

Subject:

Math

Topic:

Real Variables

Posting ID:

48213

OTA ID:

104119

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Real analysis Topology and sigma algebra

1). Prove that any sigma-algebra, which contains a finite number of memebers is also a topology. ( the Q in another words : to show that there exist a sequence of disjoint members of a sigma algebra which contains infinite no. of memebers). 2). Does there exist an infinite sigma-algebra which has only countably many members? ( ofcourse justify, examples, anything needed to prove any claims made in the solution)

Subject:

Math

Topic:

Real Variables

Posting ID:

48330

OTA ID:

101298

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Measurable sets and functions

1).If f: X--> C ( C is complex plane) is measurable, then prove that f^-1({0}) ( f inverse of 0 or any other point) is a measurable set in X. 2). If E is measurable set in X and if X_E ( x) = { 1 if x is in E, 0 if x is not in E} then X_E is a measurable function. Now I want you to prove the other direction, that is, I want you to show that If X_E(x) is measurable, then E is measurable. I believe that X_E(x) here is the characteristic function ( not sure about the name, but it is defined above). ( these 2 questions for REAL ANALYSIS class, graduate level).

Subject:

Math

Topic:

Real Variables

Posting ID:

48745

OTA ID:

101298

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Measurable functions

Suppose u(x) : X--> R v(x) : X --> R Both u(x) and v(x) are measurable Let f(x) : x --> R^2 f(x) = (u(x), v(x) ) Then f (x) is measurable Now prove a generalization of the above. That is, prove: if u_1(x) : X--> R u_2(x): X--> R . . . . u_n(x) : X--> R u_1,...,u_n are measurable and if f(x) : X --> R^n then prove that f = ( u_1,u_2,...,u_n) is measurable Please I want a detialed proof and justify every step you make, I need this next hour if possible..thanks.

Subject:

Math

Topic:

Real Variables

Posting ID:

48784

OTA ID:

101298

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