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

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

2 Answers 2


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}

  • $\begingroup$ 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. $\endgroup$
    – peek-a-boo
    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 $\int g$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .