# Dual result of Fatou lemma

If $\{f_n\}\subset L^+$, $f\in L^+$ such that $\{f_n\}$ is dominated by $f$ where $\int f < \infty$ then $$\limsup\int f_n\leq \int \limsup f_n$$

Attempted proof - Consider the sequence $\{f - f_n\}\subset L^+$ then applying Fatou's lemma we have $$\int (\liminf(f-f_n))\leq \liminf \int (f - f_n)$$ We know that $$\int(\liminf (f - f_n)) = \int \lim_{k\rightarrow \infty} \inf_{n\geq k}(f - f_n)$$ Then from the Monotone Convergence theorem, $$\int \lim_{k\rightarrow \infty}\inf_{n\geq k}(f-f_n) = \lim_{k\rightarrow \infty}\int \inf_{n\geq k}(f - f_n)$$ Therefore we have, $$\lim_{k\rightarrow \infty}\int \inf_{n\geq k}(f - f_n) = \lim_{k\rightarrow \infty}\int \inf_{n\geq k} f - \lim_{k\rightarrow \infty}\int \inf_{n\geq k}f_n = \int f - \lim_{k\rightarrow \infty}\int \sup_{n\geq k}f_n$$

I am not sure if I am on the right track or where to go from here. Just some hints should help me enough.

Attempted proof #2 - Consider the sequence $\{f - f_n\}\subset L^+$ then applying Fatou's lemma we have \begin{align*} &\int (\liminf(f-f_n))\leq \liminf \int (f - f_n)\\ \Leftrightarrow &\int\liminf f - \int \liminf f_n \leq \liminf\int f - \liminf\int f_n\\ \Leftrightarrow &\int f - \int \limsup f_n \leq \int f - \limsup\int f_n \end{align*} Rearranging and cancelling out $\int f$ (which is possible since $\int f <\infty$) then we have $$\limsup\int f_n\leq \int \limsup f_n$$

• $\sup_{n\geqslant k}f-f_n$ is decreasing with $k$, so monotone convergence would not apply. Jun 24, 2016 at 23:33
• bummer, how should I solve this then? Jun 24, 2016 at 23:34
• Apply Fatou's lemma to the sequence $\{f-f_n\}$. Jun 24, 2016 at 23:39
• @carmichael561 ok I re-edited it could you provide another hint? Jun 24, 2016 at 23:45
• For any sequence $\{f_n\}$, $\liminf(-f_n)=-\limsup f_n$. So use this and the linearity of the integral, then cancel $\int f$ and rearrange. Jun 25, 2016 at 0:01

You may try proving and using the following dual to the monotone convergence theorem:

Suppose $\{g_n\}\subseteq L^+$, $g_n$ decreases pointwise to $g$ , and $g_1$ is integrable. Then, $\int g=\lim\int g_n$.

This is Exercise 2.15 in Folland (1999, p.52).

Congratulations! Your proof 2 is more direct, and it is correct.

If $\{f_n\}\subset L^+$, $f\in L^+$ such that $\{f_n\}$ is dominated by $f$ where $\int f < \infty$ then $$\limsup\int f_n\leq \int \limsup f_n$$

Proof 2 - Consider the sequence $\{f - f_n\}_n\subset L^+$ then applying Fatou's lemma we have \begin{align*} &\int (\liminf(f-f_n))\leq \liminf \int (f - f_n)\\ \Leftrightarrow &\int\liminf f - \int \liminf f_n \leq \liminf\int f - \liminf\int f_n\\ \Leftrightarrow &\int f - \int \limsup f_n \leq \int f - \limsup\int f_n \end{align*} Rearranging and cancelling out $\int f$ (which is possible since $\int f <\infty$) then we have $$\limsup\int f_n\leq \int \limsup f_n$$

Now let us look to your proof 1. The appproach you took here is viable, but there are some adjustments need.

Proof 1 - Consider the sequence $\{f - f_n\}\subset L^+$ then applying Fatou's lemma we have $$\int \liminf(f-f_n)\leq \liminf \int (f - f_n)$$ We know that $$\int\liminf (f - f_n) = \int \lim_{k\rightarrow \infty} \inf_{n\geq k}(f - f_n)$$ Since $\{\inf_{n\geq k}(f - f_n)\}_k$ is a non_decreasing sequence of non-negative functions in $L^+$ and $\{\inf_{n\geq k}(f - f_n)\}_k \to \lim_{k\rightarrow \infty} \inf_{n\geq k}(f - f_n)$, we have, from the Monotone Convergence theorem, $$\int \lim_{k\rightarrow \infty}\inf_{n\geq k}(f-f_n) = \lim_{k\rightarrow \infty}\int \inf_{n\geq k}(f - f_n) \tag{1}$$

Now, since for all $k$ and $n\geq k$, $\inf_{n\geq k}(f - f_n)\leq f_n$, we have that, for all $k$ and $n\geq k$,
$$\int \inf_{n\geq k}(f - f_n)\leq \int (f-f_n)$$ So we have, for all $k$,
$$\int \inf_{n\geq k}(f - f_n)\leq \inf_{n\geq k}\int(f- f_n)$$ So we have $$\lim_{k\rightarrow \infty}\int \inf_{n\geq k}(f - f_n)\leq \lim_{k\rightarrow \infty}\inf_{n\geq k}\int(f- f_n) \tag {2}$$ From $(1)$ and $(2)$, we have $$\int \lim_{k\rightarrow \infty}\inf_{n\geq k}(f-f_n) \leq \lim_{k\rightarrow \infty}\inf_{n\geq k}\int(f- f_n) \tag{3}$$

Therefore we have, \begin{align*} \int f- \int\lim_{k\rightarrow \infty}\sup_{n\geq k} f_n&=\int (f- \lim_{k\rightarrow \infty}\sup_{n\geq k}f_n)=\int \lim_{k\rightarrow \infty}\inf_{n\geq k}(f-f_n)\leq \lim_{k\rightarrow \infty}\inf_{n\geq k} \int (f - f_n)= \\ &=\lim_{k\rightarrow \infty}\inf_{n\geq k} \left ( \int f - \int f_n \right)= \int f - \lim_{k\rightarrow \infty}\sup_{n\geq k}\int f_n\\ \end{align*}

Since $\int f<+\infty$, we can conclude that $$\lim_{k\rightarrow \infty}\sup_{n\geq k}\int f_n \leq \int\lim_{k\rightarrow \infty}\sup_{n\geq k} f_n$$ that is to say: $$\limsup \int f_n \leq \int\limsup f_n$$

Remark: Please note that your attempted proof 1, using Monotone convergence theorem is viable, but when we write down the details, we can see that we proving again Fatou's Lemma as part of proof 1. Please note that $(3)$ is essently Fatou's Lemma applied to $\{f-f_n\}_n$. And the rest of proof 1 after $(3)$ is essentially your proof 2. That is why Proof 2 is more concise, more direct and more elegant.