Saturday, 28 March 2015

Prove that mu is a measure under several conditions.



Suppose that F is a σ-algebra on a set X and μ:F[0,] satisfies the conditions:




  1. μ()=0.

  2. For every pair A and B of disjoint sets in F, μ(AB)=μ(A)+μ(B).


  3. For every decreasing sequence {En} in F (that is En+1En for all n) such that n=1En=, we have lim.



Prove that \mu is a measure on {\cal F}.



Here's my attempt:



Proof.



Let \{E_n\} be a countably infinite collection of sets such that E_i \cap E_j =\emptyset for all i,j. Write

E = \bigcup_{n=1}^\infty{E_n}
and let
F_n = E \setminus\bigcup_{k=1}^n{E_k}.
for n\geq 1.
Then we have
F_{n+1}= E \setminus\bigcup_{k=1}^{n+1}{E_k} \subseteq E \setminus\bigcup_{k=1}^n{E_k}=F_n
and
\bigcap_{n=1}^\infty{F_n} = \emptyset.

Hence, by applying condition (2), we have
\begin{align*}\mu\left(F_n\right) & =\mu\left(E \setminus\bigcup_{k=1}^n{E_k}\right)\\ & = \mu(E) - \mu\left(\bigcup_{k=1}^n{E_k}\right)\\ & = \mu(E) - \sum_{k=1}^n{E_k} \end{align*}
and the above holds for all n\in \mathbb{N}. Thus, applying condition (3), we have
\begin{align*}\mu(E) & = \lim_{n\to \infty}\mu(F_n) + \lim_{n\to \infty}\sum_{k=1}^n{E_k}\\ & = \sum_{k=1}^\infty{E_k}. \end{align*}
This shows that \mu is a measure.



Answer



Your proof is correct.



There are some minor typos in some places, where you wrote E_k instead of \mu(E_k), but I am sure you meant the right thing.


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