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MathematicsIsn't the statement of the Fatou's lemma somewhat problematic?
[+1] [2] Leo
[2015-04-28 18:38:53]
[ measure-theory probability-theory lebesgue-integral lebesgue-measure limsup-and-liminf ]
[ https://math.stackexchange.com/questions/1256325/isnt-the-statement-of-the-fatous-lemma-somewhat-problematic ]

My lecture notes define $\int f := \int f^+ - \int f^-$ provided both $\int f^{\pm}$ are finite. And then the Fatou's lemma is stated in the following way:

Let $f_n$ be a sequence of integrable functions and $g$ an integrable function s.t. $f_n \geq g$. Then $$\int \liminf f_n \leq \liminf \int f_n.$$

It occurred to be that given the definition, we might have a sequence that satisfies the lemma's assumptions yet $\int \liminf f_n$ is undefined.

E.g. $\Omega:= [-1,1]$ with Lebesgue measure and $$f_n:=n \mathbf 1_{[0,1]}-\mathbf 1_{[-1,0)}.$$ Then $$\liminf f_n = \infty \mathbf 1_{[0,1]}-\mathbf 1_{[-1,0)}$$ and thus $\int (\liminf f_n)^+ = \infty$ making $\int \liminf f_n$ undefined.

Does it indeed count as inconsistency? Is there a good way to repair it? (Except demanding that at least one of $\int f^{\pm}$ is finite in the definition of integrability.)

Should not Fatou by valid only for a sequence of non-negative measurable function ? What about that $g$ ? This is not the standard Fatou I know - Alonso Delfín
It is a simple and useful extension. Apply the "standard" fatou to $f_n - g \ge 0$. - Umberto P.
[+2] [2015-04-28 18:44:51] Umberto P. [ACCEPTED]

Sometimes a distinction is drawn between integrable and summable functions. If $\int f^{\pm}$ are both finite, then $\int f = \int f^+ - \int f^-$.

If only one of $\int f^\pm$ is finite, the function is summable. In case $\int f^+ = \infty$ you define $\int f = \infty$ and in case $\int f^- = \infty$ you define $\int f = -\infty$.

Summable functions don't form a vector space, but you can still make sense of things like Fatou's lemma even though the integrals involved could be infinite.


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[+1] [2015-04-28 18:42:32] Matt Samuel

Notice that the integral would only be undefined if both the positive and the negative parts are not integrable. The lower bound on the sequence guarantees that this won't happen, but there is no guarantee that the limit will be integrable. In the case that it is not you will notice that the inequality holds anyway, if you allow infinities.


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