# Isn't the statement of the Fatou's 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 Apr 28 '15 at 18:43
• It is a simple and useful extension. Apply the "standard" fatou to $f_n - g \ge 0$. – Umberto P. Apr 28 '15 at 18:45

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 $\int f = -\infty$.