1. Two Real Analysis Q's

Hi!

How would one show the following:

1. If $\displaystyle \mu(X)<\infty$, $\displaystyle \mu$ is a measure iff $\displaystyle \mu$ is continuous from above (i.e. if $\displaystyle \{E_j\}_1^\infty \subset M\subset P(X), E_1\supset E_2\supset \ldots$ and $\displaystyle \mu(E_1)<\infty$ then $\displaystyle \mu(\cub_1^{\infty}E_j)=lim_{j\rightarrow\infty}(E _j)$).

($\displaystyle \Rightarrow$) I managed this direction, it's the ($\displaystyle \Leftarrow$) that I need help with.

2. If $\displaystyle \mu$ is semifinite and $\displaystyle \mu(E)=\infty$, for $\displaystyle C>0$ there exists $\displaystyle F\subset E$ with $\displaystyle C<\mu(F)<\infty$.

Ok, so by def of semifinite we know that ther exists an $\displaystyle F'\subset E$ such that $\displaystyle 0<\mu(F')<\infty$. Could I just set $\displaystyle F=F'$?

2. On (1), it seems reasonable to check the conditions in the definition of a measure. However, I'm not exactly sure how to help you, because I don't know what kinds of assumptions you are putting on the function $\displaystyle \mu$. Are you taking $\displaystyle \mu(X)<\infty$? Is $\displaystyle \mu$ assumed to be nonnegative?

On (2), you're not quite done. You need to show that given $\displaystyle C>0$, you can produce $\displaystyle F$ so that its measure is greater than $\displaystyle C$. Try using finite / countable additivity with the set $\displaystyle F'$ that you found.

3. On (1): Yes, I'm assuming that $\displaystyle \mu$ is nonnegative and that $\displaystyle \mu(X)=\infty$ not $\displaystyle <\infty$.
$\displaystyle (X,M)$ is the meausrable space

On (2): I'm not quite sure what you mean. In order to use finite/countable additivity I need disjoint sets, and in order to form these I need at least two sets... however I only have $\displaystyle F'$.

4. Originally Posted by mgarson
On (1): Yes, I'm assuming that $\displaystyle \mu$ is nonnegative and that $\displaystyle \mu(X)=\infty$ not $\displaystyle <\infty$.
$\displaystyle (X,M)$ is the meausrable space
The first thing that we can try to show is that the measure of the empty set is zero. But I don't see how that follows only from the condition in the problem. In particular, I don't see how to eliminate the possibility that $\displaystyle \mu(\emptyset)=\infty$.

Originally Posted by mgarson
On (2): I'm not quite sure what you mean. In order to use finite/countable additivity I need disjoint sets, and in order to form these I need at least two sets... however I only have $\displaystyle F'$.
You can try setting that first set to be $\displaystyle F_1$. Then consider $\displaystyle X\backslash F_1$. That will be a set with infinite measure, so there exists a subset $\displaystyle F_2$ of finite measure. Then consider $\displaystyle X\backslash (F_1\bigcup F_2)$. That will be a set of infinite measure, so there exists a subset $\displaystyle F_3$ of finite measure. Keep iterating that procedure to get a sequence of (disjoint) sets. The real work is to show that $\displaystyle \sum \mu(F_n)$ does not converge (which I have not worked out).