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Problem: Let $(X, \mathcal{A}, \mu)$ be a measure space. Let $f: X \to [0, \infty)$ be measurable. Then define the set $$A_f = \left\{g \in L^1 (\mu)\ |\ |g| \leq f\mbox{ a.e.} \right\}.$$ Prove the following:

  1. $A_f$ is closed.
  2. If $A_f$ is norm bounded, then $f \in L ^1 (\mu)$.

I proved the first item, but I'm having trouble with the second. I have a feeling that I have to use Lebesgue's dominated convergence theorem, but I have no clue how to set it up. This is probably really easy, but I just couldn't find the solution for the last couple hours. Any help is appreciated.

P.S. This question is from an old exam, it's not a homework question or anything like that.

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  • $\begingroup$ Note $f \in A_f$, since $|f| \le f$. $\endgroup$
    – GEdgar
    Commented Jun 13 at 14:12
  • $\begingroup$ @GEdgar This is what I have to prove, yes. $\endgroup$
    – the_dude
    Commented Jun 13 at 15:51
  • $\begingroup$ @the_dude Hi sorry if I am bothering you . How did you show this set is closed ? I would have used the fact $g_n \rightarrow g$ in $L_1$ implies $g_(n_k) \rightarrow g$ a e so passing with limit in $k$ i should have the closure . Please let me know $\endgroup$ Commented Jun 17 at 23:35
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    $\begingroup$ @Dsrksidemath Sorry I have been inactive for a couple days, so I didn't see your question. I did it the same way as you have: take the sequence $g_n$ in $A_f$ that converges to some $g \in L^1 (\mu)$. Then by Riesz-Fischer, there exists a subsequence $g_{n_k} \to g$ almost everywhere. Define $B_0$ as the set where $g_{n_k} \to g$ pointwise and $B_k$ as the set where $|g_{n_k}| \leq f$. Finally, define $C := \bigcap_{k = 0} ^\infty B_k$ and then $\mu(C^{\mathsf{c}}) = 0$ and $|g| \leq f$ on $C$, so $g \in A_f$. $\endgroup$
    – the_dude
    Commented Jun 23 at 17:22

1 Answer 1

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I think that we have to assume the measure space to be $\sigma$-finite. Otherwise, if we are in a space where $\mu(A)=\infty$ if $A$ is non-empty and $\mu(\emptyset)=0$, then taking $f$ constant equal to one would give $A_f$ reduced to the constant map equal to $0$ but $f$ is not in $L^1$.

Let $(X_n)$ be an increasing sequence of elements of $\mathcal A$ such that $X=\bigcup_{n\geqslant 1}X_n$ and $\mu(X_n)$ is finite. For each integer $n$, define the function $f_n\colon x\mapsto f(x)\mathbf{1}_{\{ f(x) \leqslant n\}}\mathbf{1}_{X_n}(x)$. Then $f_n\in A_f$ and $f_n\in L^1$ and it follows that $\lVert f_n\rVert_1\leqslant \sup_{g\in A_f}\lVert g\rVert_1<\infty$. Conclude by monotone convergence.

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