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Let μ be a measure in Ω and A be afixed set in Ω .Then show that λ defined as λ(E)=μ(A∩E) is a measure on Ω for E in Ω. How to prove this?

Shammu
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  • By proving that $\lambda$ has the characteristic properties of a measure. I reckon you are familiar with them. If not then find them e.g. by googling. – drhab Dec 10 '17 at 15:01
  • How to prove that..plz explain – Shammu Dec 10 '17 at 15:02
  • $\lambda(E)=\mu(A\cap E)\geq0$ for instance. A measure takes nonnegative values. What is your problem? Can you prove that $\lambda(\varnothing)=0$ and $\lambda(\bigcup_{n=1}^{\infty} (E_n)=\sum_{n=1}^{\infty}\lambda(E_n)$ if the $E_i$ are disjoint? Then you are ready. – drhab Dec 10 '17 at 15:08
  • You must prove that $\lambda$ is a measure. By fixed $A$,$E$ $\mu(A\cap E)$ is just some element of $[0,\infty]$. – drhab Dec 10 '17 at 15:12

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Let it be that $\mu$ is a measure on measurable space $(\Omega,\mathcal A)$.

That means that $\mu$ is a function on $\mathcal A$ with specific properties (see the link).

Now for some fixed $A$ we prescribe the function $\lambda$ on $\mathcal A$ by $E\mapsto\mu(A\cap E)$.

To be shown is that also $\lambda$ has the characteristic properties that you can find in the link.

One of them is that $\lambda(\varnothing)=0$.

This follows from $\lambda(\varnothing)=\mu(A\cap\varnothing)=\mu(\varnothing)=0$.

I leave the rest to you.

drhab
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  • Examine the following sequences for convergence if convergent,derive the limit. (1)An=[-1-(1/n),1-(1/2n)] (2)A2n=(0,1/2n). , A2n+1=(-1,1/2n+1) – Shammu Dec 10 '17 at 16:10
  • There is no connection at all between your comment and my answer to your question. – drhab Dec 11 '17 at 13:35