Home > Measure Theory, Real Analysis > Characterization of sigma-finite measure spaces

Characterization of sigma-finite measure spaces

Let {(X,\mathcal{E})} be a measure space and {\mu} be a positive measure on it; prove that {L^1(X,\mu)} contains a strictly positive function if and only if {X} is {\sigma}-finite with respect to {\mu}.

Ambrosio et al., Functions of Bounded Variation and Free Discontinuity Problems, Ex 1.5

Proof: Suppose {L^1(X,\mu)} contains a strictly positive function {f}. Then it is easy to see that {A_n=\{ x \in X : f(x) \geq \frac{1}{n}\}} forms a partition of {X} into sets of finite measure.

Conversely, suppose that {X} is {\sigma}-finite and that {A_n,\ n \geq 0} are the finite measure sets which partition {X}. Define

\displaystyle f(x)=\sum_{n=0}^\infty \frac{1}{2^n}\cdot \frac{\chi_{A_n}}{\mu(A_n)+1}

and note that {f} is strictly positive, it is well defined since the series is always convergent and moreover, its integral is finite. This may seem as an abstract guess, but, in fact, is very natural.

Here are a few steps I took into finding this example:

  • (i) Since the only hypothesis we have is that {X} is {\sigma}-finite, we must use the finite measure sets {(A_n)} which decompose {X} to construct a function, and how can we do that if not by using their characteristic functions?
  • (ii) Since the desired function must be strictly positive we could think of our function as a series of {\chi_{A_n}}, but we quickly see that it may happen that both {f(x)=\sum_n \chi_{A_n}} and {f(x)=\sum_n \frac{1}{2^n}\chi_{A_n}} may not have finite integrals, since the measures of {A_n} may be finite, but not small enough.
  • (iii) To fix this divide every term of the sum by a number bigger than {\mu(A_n)} and voilà.
Categories: Measure Theory, Real Analysis Tags:
  1. No comments yet.
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: