lebesgue sets and compact sets problem

If A is lebesgue measurable sets in R^n, bounded, then there is a compact set K_epsilon and an open set for every epsilon > 0 V_epsilon such that K_epsilon is subset of A and A is a subset of V_epsilon and for m(A-K) < epsilon m(V-K) < epsilon

lebesgue measurable set

Let A be a set in R^n, we denote by A + x_o a parallel shift of A by x_o to A + x_o, A + x_o = { x : x = y + x_o, y in A}. Now, if A is a lebesgue measurable then show that 1). x_o + A is also lebesgue measurable 2). m(A) = m(x_o + A) Can someone check my answer and tell me if it is correct or not? My work: since A is a subset of R^n then A^c ( A compliment) = R^n - A and therefore A^c union A = R^n and (A + x_o ) union (A^c + x_o) = R^n Since A+x_o in R^n and (A+x_o)^c = A^c + x_o ( to show (A + x_o)^c = A^c + x_o : let z be element of (A+x_o)^c then z doesn't belong to A + x_o, since x_o is a point then z doesn't belong to x_o, and z doesn't belong to A so z belong... click for more

Properties of integrals of SIMPLE FUNCTIONS

1).If A is a subset of B, A,B in m ( measurable sets) then show that integral (region A) s dM =< integral ( region B) s dM Where s here is a simple non-negative measurable function. ( Please don't confuse this with bounded measurable functions, I need the proof for SIMPLE functions). 2). If E are measurable, X_E is the charachteristic function, s is as defined in 1, then show that integral (region E) s d M = integral (region X) X_E s D M. *****PLEASE MAKE SURE IN YOUR PROOF TO SHOW THAT THE PRODUCT OF SIMPLE FUNCTION AND THE CHARACHTERISTIC FUNCTION IS AGAIN A SIMPLE FUNCTION.****** AGAIN ALL THESE RESULTS I WANT TO PROVE ARE FOR SIMPLE FUNCTIONS. I WANT THIS FOR SIMP... click for more

Polynomial

(See attached file for full problem description with proper symbols) --- Let and for (a) Use integration by parts to show that in for . Deduce that for (b) Compute for and verify that ---

Lower measures

True or False problem. m_* (A) = Sup sum_i | M_i| ( U M_i is subset of A) Where m_* is the inner measure M_i doesn't equal M_j for i doesn't equal j ( i.e, they are disjoint) Prove it or show a counterexample and explain it to show how the equality doesn't hold.