if $\nu(E) = \sum_{k=1}^{n} c_{k} \mu(E \bigcap E_{k}).$ Show that $\nu$ is a measure on $(X, \mathfrak{M})$

Question

Let $(X, \mathfrak{M}, \mu)$ be a finite measure space, let $\{E_{k}\}_{k=1}^{n}$ a collection of measurable sets, and $\{ c_{k}\}_{k=1}^n$ a collection of real numbers. For $E \in \mathfrak{M}$ define $$\nu(E) = \sum_{k=1}^{n} c_{k} \mu(E \bigcap E_{k}).$$ Show that $\nu$ is a measure on $(X, \mathfrak{M})$ that is absolutely continuous with respect to $\mu$ and find its Radon-Nikodym derivative $d\nu / d\mu$

How can I show the countable additivity of $\nu$?

My trial:

$\nu(\bigcup_{i = 1}^{\infty} F_{i}) = \sum_{k=1}^{n} c_{k} \mu(\bigcup_{i = 1}^{\infty} F_{i} \bigcap E_{k}) = \sum_{k=1}^{n} c_{k} \mu(\bigcup_{i = 1}^{\infty} (F_{i} \bigcap E_{k})) = \sum_{k=1}^{n} c_{k} \sum_{i=1}^{\infty} \mu (F_{i} \bigcap E_{k}) = \sum_{i=1}^{\infty} \sum_{k=1}^{n} c_{k} \mu(F_{i} \bigcap E_{k}) = \sum_{i=1}^{\infty} \nu (F_{i})$

score 3 · Accepted Answer · answered Jan 27 '20 at 10:08

3

$E \to \mu(E\cap E_k)$ is countably additive for each $k$.

If $\mu$ and $\nu$ are countable additive so is $a\mu+b\nu$ for any constants $a$ and $b$. [This is immediate from definition]. Now use induction on $n$.

answered Jan 27 '20 at 10:08

Kavi Rama Murthy

311,013

and why the given measure is nonnegative? the $c_{k}$ are real numbers. – Jan 27 '20 at 17:13
@Mathstupid There are what are called real measures, complex measures, vector measures etc. This question does not ask you to prove that $\nu$ is non-negative and you cannot prove that. All you have to prove is countable additivity. – Kavi Rama Murthy Jan 27 '20 at 23:12
Is not the countable additivity requires an assumption that $\nu(\bigcup_{k=1}^{\infty} E) = \sum_{k=1}^{\infty} \nu (E_{k})$? – Emptymind Jan 28 '20 at 00:04
Yes, but this equation can hold for set functions which are not non-negative. The equation means that the series of real numbers on the right side converges and its sum is equal to the left side. – Kavi Rama Murthy Jan 28 '20 at 00:09
No, I am speaking about your way of proving countable additivity .... not the comment of Mathstupid .... I will edit my question to include how I want to prove it .... and just please check it for me. – Emptymind Jan 28 '20 at 00:15
I can not see any series on the r.h.s .... do you mean a finite series? – Emptymind Jan 28 '20 at 00:17
@Emptymind Isn't $ \sum\limits_{k=1}^{\infty} \nu (E_k)$ a series? – Kavi Rama Murthy Jan 28 '20 at 00:21
yess it is .... by the way I included my trial. – Emptymind Jan 28 '20 at 00:29
@Emptymind Yes, you got the proof now. You can always exchange the sums when one of them is a finite sum. – Kavi Rama Murthy Jan 28 '20 at 00:30
I did not use any induction – Emptymind Jan 28 '20 at 00:33
and is exchanging the addition sign (one is infinite and one finite ) ok? – Emptymind Jan 28 '20 at 00:34

if $\nu(E) = \sum_{k=1}^{n} c_{k} \mu(E \bigcap E_{k}).$ Show that $\nu$ is a measure on $(X, \mathfrak{M})$

1 Answers1