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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.
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