# Understanding the assumptions in the Reverse Fatou's Lemma

Fatou's Lemma says the following:

If $(f_n)$ is a sequence of extended real-valued, nonnegative, measurable functions defined on a measure space $\left(\mathbf{X},\mathcal{X},\mu\right)$, then $$\int\lim\inf f_n d\mu \leq \lim\inf \int f_n d\mu.$$

In the statement of the Reverse Fatou's Lemma there's an addtional requirement that the given sequence be dominated by an integrable function. I'm interested in understanding what breaks down if this condition is not satisfied. For the sake of clarity and notation, here's the statement of the Reverse Fatou's Lemma:

Let $(f_n)$ be a sequence of extended real-valued functions defined on a measure space $\left(\mathbf{X},\mathcal{X},\mu\right)$. If there exists an integrable function $g$ on $\mathbf{X}$ such that $f_n \leq g$ for all $n$, then $$\lim\sup\int f_n d\mu \leq \int\lim\sup f_n d\mu.$$

Again, I'm curious to know what happens if this additional condition that the sequence be dominated is not satisfied. In the proofs that I've seen of the Reverse Fatou's Lemma they've all taken advantage of the fact that the functions are dominated, but I just don't see why there can't be a proof of the inequality that doesn't use this assumption.

My interest was further piqued by the following problem I came across in Bartle's Elements of Integration and Lebesgue Measure:

Let $(f_n)$ be a sequence of extended real-valued, nonnegative functions defined on $\left(\mathbf{X},\mathcal{X},\mu\right)$, $f_n \to f$, and let $\int f d\mu =\lim \int f_n d\mu < \infty.$ Show that for any $E \in \mathcal{X},$ $$\int_E f d\mu =\lim \int_E f_n d\mu.$$

I was able to prove this through two applications of Fatou's Lemma and use of the nice identity $\lim\sup(-f_n) =-\lim\inf(f_n)$. But there was another proof I abandoned after I failed to prove that the Reverse Fatou's Lemma held with the given hypotheses.

Any insight is much appreciated.

For $$A>1$$, take $$f_n=n^{-1}\mathbf{1}_{[n,An)}$$ on $$\mathbb R_+$$ endowed with Lebesgue measure. Then for each $$x$$, $$f_n(x)\to 0$$, but $$\int f_n(x)d\lambda =A-1$$. One has that $$\sup_{n\geqslant 1}f_n(x)\geqslant c/x$$ for some constant $$c$$ hence an integrable function $$g$$ that dominates all the $$f_n$$ does not exist.
We can also consider an example where $$\int f_n(x)d\lambda\to \infty$$: take $$f_n=n^{-1}\mathbf{1}_{(n,n^2)}$$.
• My apologies if I'm mistaken... For convergent examples isn't this just nothing but the usual Fatou: $f_n(x):=\chi_{(n,n+1)}\to0:\quad0=\int_\mathbb{R}\lim_nf_n\mathrm{d}x\leq\lim_n\int_\mathbb{R}f_n\mathrm{d}x=1\quad(" \liminf=\lim=\limsup")$ For an honest example one needs to come up with divergent example as: $f_{1\leq k\leq n}(x):=\chi_{(\frac{k-1}{n},\frac{k}{n}]}:\quad1=\int_{[0,1]}\limsup_{n;k}f_{n;k}(x)\mathrm{d}x\nleq\limsup_{n;k}\int_{[0,1]}f_{n;k}(x)\mathrm{d}x=0\quad(" \liminf\neq\lim\neq\limsup")$ Dec 16, 2014 at 18:50