# The domination assumption in Fatou's lemma and the dominated convergence theorem.

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 $$f = -n^2 \chi_{[0, \frac{1}{n}]}$$.

• For Fatou's Lemma it is clear that $g\geq 0$ is not necessary, but it is the less strict condition since one requires $f_n\geq \color{red}{-}g$. Mar 5, 2022 at 23:22

The usual Fatou's lemma is

If $$\{\phi_n\}$$ is a sequence of measurable functions such that $$\phi_n\geq 0$$ (a.e) for all $$n$$, then $$\int\liminf \phi_n\leq \liminf\int \phi_n$$.

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):

Suppose $$\{f_n\}$$ and $$g$$ are measurable functions such that for all $$n$$ we have $$f_n\geq g$$ a.e. If $$g$$ is integrable then $$\int \liminf f_n\leq \liminf \int f_n$$.

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}

• Note btw that if you have now have $|f_n|\leq g$, with $g$ integrable, then you can use both versions of Fatou to get $\int\liminf f_n\leq \liminf \int f_n\leq \limsup \int f_n\leq \int \limsup f_n$. So, if you make the further assumption that $f_n\to f$ pointwise a.e, then this shows $\int f\leq \liminf \int f_n\leq \limsup \int f_n\leq \int f$, so that $\lim \int f_n$ exists and equals $\int f$. In other words, this proves the dominated convergence theorem. Mar 6, 2022 at 3:33

You have already answered the second question. For the first one the answer is YES: Suppose $$f_n \geq g$$ and $$g$$ is integrable. Then, $$f_n-g$$ is non-negative so $$\int \lim \inf (f_n-g) \leq \lim \inf \int (f_n-g)=\lim \inf \int f_n-\int g$$ so $$\int \lim \inf f_n-\int g\int [\lim \inf f_n-g] \leq \lim \inf \int f_n-\int g$$. Now cancel $$\int g$$.