# Two Real Analysis Q's

• Jan 29th 2011, 06:53 AM
mgarson
Two Real Analysis Q's
Hi!

How would one show the following:

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

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

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

Ok, so by def of semifinite we know that ther exists an $F'\subset E$ such that $0<\mu(F')<\infty$. Could I just set $F=F'$?
• Jan 29th 2011, 07:08 AM
roninpro
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 $\mu$. Are you taking $\mu(X)<\infty$? Is $\mu$ assumed to be nonnegative?

On (2), you're not quite done. You need to show that given $C>0$, you can produce $F$ so that its measure is greater than $C$. Try using finite / countable additivity with the set $F'$ that you found.
• Jan 29th 2011, 12:16 PM
mgarson
On (1): Yes, I'm assuming that $\mu$ is nonnegative and that $\mu(X)=\infty$ not $<\infty$.
$(X,M)$ 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 $F'$.
• Jan 29th 2011, 02:15 PM
roninpro
Quote:

Originally Posted by mgarson
On (1): Yes, I'm assuming that $\mu$ is nonnegative and that $\mu(X)=\infty$ not $<\infty$.
$(X,M)$ 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 $\mu(\emptyset)=\infty$.

Quote:

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 $F'$.

You can try setting that first set to be $F_1$. Then consider $X\backslash F_1$. That will be a set with infinite measure, so there exists a subset $F_2$ of finite measure. Then consider $X\backslash (F_1\bigcup F_2)$. That will be a set of infinite measure, so there exists a subset $F_3$ of finite measure. Keep iterating that procedure to get a sequence of (disjoint) sets. The real work is to show that $\sum \mu(F_n)$ does not converge (which I have not worked out).