Find a function $\psi$ s.t. $\forall f\in\mathcal{F},f\le\psi$ a.e. , where $\mathcal{F}$ is a given subset in $L^{1}(\mathbb{R})$

Question

Prove that if $\mathcal{F}$ is a nonempty subset in $L^{1}(\mathbb{R})$ s.t.

(1) if $f,g\in\mathcal{F}$, $h(x) = \max(f(x),g(x))$, then $h\in\mathcal{F}$,

(2)$M: = \sup_{f\in\mathcal{F}}(\int_{R}f(x)dx)<\infty$,

then there exists a function $\psi$ s.t.

(a) $\forall f\in\mathcal{F},f\le\psi$ a.e. ,

(b)$\int_{R}\psi(x)dx=M$.

I have tried to prove that there exists a function $\psi\in\mathcal{F}$ s.t. $\forall f\in\mathcal{F},f\le\psi$ a.e. which seems like the conclusion of Zorn's lemma. However the relation "$f\le g$ a.e." isn't a strict partial order because $f\le g$ a.e. and $g\le f$ a.e. can only derive that $f=g$ a.e. instead of $f=g$. Maybe we need to solve the problem using functional analysis but I haven't got a good idea.

Answer 1

See this page for a generalization of your result, but it's really quite simple.

Let $f_n\in \mathcal F$ be any sequence of functions such that $\int_{\mathbb R} f_n(x)dx \to M$. By taking $g_n = \max\{f_1,f_2,\ldots,f_n\}$, we know that $g_n \in \mathcal F$ and $M \geq \int_{\mathbb R} g_n(x)dx \geq \int_{\mathbb R} f_n(x)dx$. By the squeeze theorem, $\int_{\mathbb R} g_n(x)dx \to M$.

As $g_n$ is a monotonically increasing sequence of functions, we know that $g_n(x)$ for all $x \in \mathbb R$ has a pointwise limit $g(x)$ that is either finite or positive infinite (every monotonic increasing sequence converges or diverges to positive infinity). By properties of measurable functions, $g(x)$ is measurable , and $\int_{\mathbb R} g_n(x)dx \to \int_{\mathbb R}g(x)dx$ by the monotone convergence theorem. It follows that $\int_{\mathbb R} g(x)dx = M$ and that $g$ is finite a.s.

We claim that $g$ satisfies condition (a) (i.e. that it is the greatest element of $\mathcal F$). To prove this, let $g'$ be any other function such that $g'>g$ on a set $S$ of positive measure $|S|>0$. Then, $\max\{g',g\} > g$ on $S$. A standard consequence of this inequality is that for some $n>1$, the set $\{\max\{g',g\} > g + \frac 1n\}=S_n$ has positive measure. We'll assume that $S_n$ has finite measure as well by truncating it if necessary.

Finally, $$ \int_{\mathbb R} \max\{g'(x),g(x)\} dx > \int_{S_n^c} g(x)dx + \int_{S_n} g(x)dx+\frac{|S_n|}{n} > M+\frac{|S_n|}{n}. $$ However, $\max\{g',g\} \in \mathcal F$ has integral greater than $M$, a contradiction. It follows that no such $g'$ exists. Consequently, $g\geq g'$ a.s. for every $g' \in \mathcal F$, which is condition (a), as desired.

Zorn's lemma was not required because we got the maximal element by explicit construction as the limit of some subsequence, rather than through the abstract machinery of general lattices.

Answer 2

Assu Let $(f_n:n\in\mathbb{N})$ be a sequence in $\mathcal{F}$ so that $\int f_n\xrightarrow{n\rightarrow\infty} M$. As $\mathcal{F}$ is closed under finite maxima, we may assume without loss of generality that $f_n\leq f_{n+1}$ (substitute $f_n$ by $\max(f_1,\ldots, f_{n-1})$ is necessary). Since $0\leq f_n-f_1$, by monotone convergence $g=\lim_nf_n$ satisfies $$\int g=\int f_1+\lim_n\int (f_n- f_1)=M$$ For any $f\in\mathcal{F}$, $f\leq \max(f,f_n)\in \mathcal{F}$, and so $\int \max(f_n,f)\leq M=\int g$.

Another application of monotone convergence ($\phi_n=\max(f_n,f)\leq \phi_n$ and so, $\phi_n-\phi_1\geq0$) implies that $$M=\int g\leq \int\max(\phi,f)=\lim_n\int\max(\phi_n,f)\leq M=\int g$$ Since $g\leq \max(f, g)$, $\max(f, g)=g$ a.s.. That is $f\leq g$ a.s.

The function $g$ is the object the OP is looking for.

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Comment: A similar idea can be used to prove a slightly stronger result: Suppose $\mathcal{F}\subset L_p(X,\mathscr{B},\mu)$, $1\leq p<\infty$, and that there is $g\in L_p(\mu)$ such that $f\leq g$ ($\mu$-a.s.). Then there is $\phi\in L_p(\mu)$ such that

• $f\leq \phi$ for all $f\in \mathcal{F}$,
• and if $h\in L_p(\mu)$ is any other function such that $f\leq h$ for all $f\in\mathcal{F}$, then $\phi\leq h$.

$\phi$ is the $L_p$-supremum of $\mathcal{F}$.