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Question

I am trying to prove the Reverse Fatou's lemma but I can't seem to get it. The statement is the following:

Suppose that $(f_n)_{n \in \mathbb{N}}$ is a sequence of measurable functions and $g$ an integrable function such that $f_n \leq g$ for all $n \in \mathbb{N}$. Then, $\limsup _{n \rightarrow \infty} \mu (f_n) = \mu ( \limsup _{n \rightarrow \infty} f_n)$ where $\mu$ is the integral on a specified measure space.

Attempt

We have a sequence $\lbrace f_k \rbrace$ in $\mathbb R$ and $E\subset \mathbb R$. We know that $\limsup f_k = \lim\limits_{j\rightarrow \infty} g_j$ where $g_j = \sup\limits_{k\geq j } f_k$. Thus we get that $$f_k \leq g_j \Rightarrow \int_E f_k \leq \int_E g_j \implies \sup\limits_{k\geq j }\int_E f_k \leq \int_E g_j $$ Taking the limit of both sides yields $$\lim\limits_{j\rightarrow \infty}\sup\limits_{k\geq j }\int_E f_k \leq \lim\limits_{j\rightarrow \infty}\int_E g_j $$ Which is equivalent to $$\limsup\limits_{j\rightarrow \infty}\int_E f_k \leq \liminf\limits_{j\rightarrow \infty}\int_E g_j $$

This is where I get stuck. I want to use Fatou's lemma but it won't work in this case. Is there a better way to prove this?

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    $\begingroup$ If the $f_k$ are dominated by an integrable function, so are the $g_j$, and the monotone convergence theorem does the rest. If the $f_k$ aren't dominated by an integrable function, $$\limsup \int f_k \leqslant \int \limsup f_k$$ need not hold. $\endgroup$ Oct 19, 2013 at 20:55
  • $\begingroup$ I thought the Monotone convergence theorem only applies to increasing functions, which we do not have. Is that not true? $\endgroup$
    – rioneye
    Oct 19, 2013 at 21:00
  • $\begingroup$ If you have a non-increasing sequence $g_j$ that is dominated by $h$, where $h$is integrable, consider $d_j = h - g_j$. (By the way, you can also directly apply Fatou's lemma to $h - f_k$ if $h$ dominates the $f_k$.) $\endgroup$ Oct 19, 2013 at 21:06
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    $\begingroup$ Then let $h_k = f - f_k$. Fatou: $\int \liminf h_k \leqslant \liminf \int h_k$. Expanding $h_k$: $\int (f - \limsup f_k) \leqslant \int f - \limsup \int f_k$, or subtracting the finite quantity $\int f$, $-\int \limsup f_k \leqslant -\limsup \int f_k \iff \limsup \int f_k \leqslant \int \limsup f_k$. $\endgroup$ Oct 19, 2013 at 22:48
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    $\begingroup$ No. I'm saying $\liminf (-f_k) = - \limsup f_k$. $\endgroup$ Oct 19, 2013 at 22:54

1 Answer 1

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As $(f_n)_{n \in \mathbb{N}}$ and $g$ are both measurable, we know that $(g-f_n)$ is also measurable.

Therefore by Fatou's Lemma

$$\mu (\liminf _{n \rightarrow \infty} (g-f_n)) \leq \liminf _{n \rightarrow \infty}\mu(g-f_n) \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space(1)$$

As the function $g$ is independent of $n$, we can rewrite $(1)$ as the following (by linearity of the integral)

$$ \mu (g) + \mu(\liminf _{n \rightarrow \infty} (-f_n)) \leq \mu(g)+ \liminf _{n \rightarrow \infty} (-\mu(f_n))\space \space \space \space \space \space \space \space \space (2)$$

We can immediately observe that $\mu (g)$ is present on both sides of the inequality which allows us to simplify to the following result:

$$ \mu(\liminf _{n \rightarrow \infty} (-f_n)) \leq \liminf _{n \rightarrow \infty} (-\mu(f_n))\space \space \space \space \space \space \space \space \space (3)$$

We now complete the proof by using the fact that $\liminf _{n\rightarrow \infty} (-f_n) = - \limsup _{n \rightarrow \infty} (f_n)$ and substituting this into $(3)$ to deduce

$$ -\mu(\limsup _{n \rightarrow \infty}(f_n)) \leq -\limsup _{n \rightarrow \infty} (\mu(f_n))\space \space \space \space \space \space \space \space \space (4)$$

And by rearranging inequality $(4)$ above, we get the final result:

$$ \fbox{$\limsup _{n \rightarrow \infty} (\mu(f_n)) \space \leq \space \mu(\limsup _{n \rightarrow \infty}(f_n))$} $$

Note: as discussed in the comments, equality does not hold here unless the limit exists and the hypotheses for monotone convergence are satisfied

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  • $\begingroup$ Thanks for nice derivation. Is the other direction obvious so that you can make it an equality ? If so, not to me. Thanks. $\endgroup$
    – mark leeds
    Jan 14, 2023 at 3:43
  • $\begingroup$ The other direction is not true in general, so we cannot make this an equality. By the Monotone Converhence Theorem if the limit exists, then we have equality. But we can’t guarantee that the limit is defined @markleeds $\endgroup$
    – FD_bfa
    Jan 14, 2023 at 10:01
  • $\begingroup$ It's much appreciated. $\endgroup$
    – mark leeds
    Jan 15, 2023 at 11:33

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