I often find these two statements representing the following two theorems.
Fatou's lemma:
i). If $f_n \geq -g$ where $g \geq 0$ is integrable, then $\int \liminf f_n \leq \liminf \int f_n$.
ii). If $f_n \leq g$ where $g \geq 0$ is integrable then $\limsup \int f_n \leq \int \limsup f_n$.
Dominated convergence theorem:
If $f_n$ converges a.e. to $f$ and $|f_n| \leq g$ with $g$ integrable, then $\lim \int f_n = \int f$.
For Fatou's lemma, is it necessary for $g \geq 0$? For example, is it enough to change the assumption to only there exists $g$ integrable such that $f_n \geq g$?
For the dominated convergence theorem, is it necessary the that $|f_n| \leq g$? Can we weaken it to only $f_n \leq g$ for integrable $g$?
My question:
Does anyone have any references to these weakened assumptions in any literature or can anyone find counterexamples? I believe for both cases, the weakened assumptions are enough…
Edit: I did some thinking and realize that $f_n \leq g$ is not enough for the dominated convergence theorem. Consider $g = 0$ and $f = -n^2 \chi_{[0, \frac{1}{n}]}$.
Best Answer
The usual Fatou's lemma is
Any other variation you encounter is easily deduced from this. How can we thus figure out correct hypotheses in "more general" cases? Well, start with any sequence $\{f_n\}$ of measurable functions and let $\gamma$ be measurable. Suppose we have $f_n\geq \gamma$ for all $n$. What assumptions should we impose on $\gamma$ so that we can conclude (that the intergals make sense and) $\int\liminf f_n\leq \liminf \int f_n$?
Well, the hypothesis $f_n\geq \gamma$ automatically suggests us to consider $\phi_n:= f_n-\gamma$. We had better assume $\gamma$ is finite a.e so that the subtraction makes sense a.e and that $f_n-\gamma\geq 0$ a.e. So, we now have a sequence $\phi_n=f_n-\gamma$ of non-negative measurable functions, so by the basic Fatou lemma, we have \begin{align} 0 \leq \int\liminf (f_n-\gamma) \leq \liminf \int (f_n-\gamma) \end{align} Using basic properties of $\liminf$, we have \begin{align} 0 \leq \int [(\liminf f_n)-\gamma]\leq \liminf \int (f_n-\gamma) \end{align}
Now, we would like to split the integral and then add $\int \gamma$ to both sides. When can we do this? Well, suppose $\gamma$ is integrable (so necessarily finite a.e). In this case, $f_n= (f_n-\gamma)+ \gamma$, so $f_n$ is the sum of a non-negative function and an integrable function. Hence, $f_n$ has a well-defined integral in $(-\infty,\infty]$ (note I'm not saying $f_n$ is integrable, because $f_n=\infty$ satisfies all these conditions but clearly isn't integrable if the measure space has strictly positive measure). Similarly, $\liminf f_n$ has a well-defined integral in $(-\infty,\infty]$. Hence, we can split things above: \begin{align} \int \liminf f_n-\int \gamma\leq \left(\liminf \int f_n\right)-\int\gamma \end{align} Since $\gamma$ is integrable, we can add $\int\gamma$ to both sides to conclude \begin{align} \int \liminf f_n\leq \liminf\int f_n. \end{align}
To summarize (with slightly different notation):
For (ii), suppose $f_n\leq g$ and $g$ is integrable. Then, $-g\leq -f_n$, so by the generalized Fatou proved above (using the fact $-g$ is also integrable), we have \begin{align} \int\liminf (-f_n)\leq \liminf\int(-f_n). \end{align} But now using the relation $\liminf (-a_n)=-\limsup(a_n)$, which is valid for sequences of numbers in general, we immediately get \begin{align} \limsup\int f_n\leq \int\limsup f_n. \end{align}