Saturday

February 28, 2015

February 28, 2015

Total # Posts: 9

**math**

Let E(1,2) be the set of all numbers in (0,1) such that there decimal representation does not contain 1 and 2. Prove that E(1,2) is lebesgue measurable and find the lebesgue measure of E(1,2). Would you please explain it step by step?
*January 17, 2009*

**lebesgue measure**

Let E(1,2) be the set of all numbers in (0,1) such that there decimal >representation does not contain 1 and 2. Prove that E(1,2) is lebesgue >measurable and find the lebesgue measure of E(1,2).
*January 17, 2009*

**pls help :(**

Let E(1,2) be the set of all numbers in (0,1) such that there decimal representation does not contain 1 and 2. Prove that E(1,2) is lebesgue measurable and find the lebesgue measure of E(1,2). Would you please explain it step by step?
*January 17, 2009*

**math**

thank you but what will ý do at the end?
*January 15, 2009*

**math**

Let S be the set of all measurable functions on [0,1]. Then the set O of all functions in S equal to 0 a.e on [0,1] makes an additive subgroup of a commutative group S. Show that S/O with the distance p(f,g)=indefinite integral(0,1)((|f(x)-g(x)|)/(1+|f(x)-g(x)))dx is a metric ...
*January 15, 2009*

**math**

Let E(1,2) be the set of all numbers in (0,1) such that their decimal representation does not contain 1 and 2. Prove that E(1,2) is Lebesgue measurable and find the Lebesgue measure of E(1,2).
*January 15, 2009*

**Lbesgue measure**

Let E(1,2) be the set of all numbers in (0,1) such that their decimal representation does not contain 1 and 2. Prove that E(1,2) is Lebesgue measurable and find the Lebesgue measure of E(1,2).
*January 15, 2009*

**smallest algebra**

becouse they couldnt help me :(
*January 15, 2009*

**smallest algebra**

Let F be a family of subsets of a set X such that empty set is the element of F. A subset A of X belongs to F1 if and only if either A is a subset of F or complement of A is the subset of F. A subset B of X belongs to F2 if and only if B is a finite intersection of sets in F1...
*January 15, 2009*

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