Closed Subset of ${ L }^{ p }$

Question

I'd like to show that the set $$ S:=\left\{ f:X\rightarrow R:\quad f\quad \mu -\mathrm{measurable},\quad \int_{ X } { f }^{ 2 }d\mu \le 1 \right\} $$ is closed in ${ L }^{ p }(d\mu )$, where $X$ is an arbitrary set and $\mu$ a finite measure on the sigma algebra $\Sigma \subset { 2 }^{ X }$.

First of all is it correct that this given set is the closed unit ball in ${ L }^{ 2 }$? I tried going with the standard definition and tried to show that our set contains all its limits points but I am having struggels to define arbitrary sequences in this space. Could someone offer a hint.

Answer 1

$d_p(f,g)=\|f-g\|_{L^p}=\left(\int_X |f-g|^p\,d\mu\right)^{1/p}$ is a distance on $L^p(X,\mu)$; this follows from the Minkowski inequality.

Your set is $B(0,1)=\{f\in L^p: d_p(f,0)\leq 1\}$ and it's a general fact that closed balls are, indeed, closed. This is just an immediate consequence of the definition of a topology induced by a metric.

Fact. In a metric space $(X,d)$, any "closed ball" $D(x_0,r)=\{x\in X:d(x,x_0)\leq r\}$ is closed in the induced topology.

Proof. Take $x\notin D(x_0,r)$. Then $d(x,x_0)>r$. Take $0<\delta<d(x,x_0)-r$. Then I claim that $B(x,\delta)\subset D(x_0,r)^c$. Indeed, if $y\in B(x,\delta)$, then $$ d(y,x_0)\geq d(x,x_0)-d(y,x)> d(x,x_0)-\delta > r. $$ This means that $D(x_0,r)^c$ is open, which is equivalent to the thesis.

Answer 2

Let $\left(f_n\right)_{n\geqslant 1}$ be a sequence in the involved set with converges in $\mathbb L^p$ to some $f$. Extract a subsequence $\left(f_{n_k}\right)_{k\geqslant 1}$ which converges almost everywhere to $f$. Using Fatou's lemma, we obtain $$ \int_Xf^2\mathrm d\mu=\int_X\liminf_{k\to +\infty} f_{n_k}^2\mathrm d\mu\leqslant \liminf_{k\to +\infty} \int_X f_{n_k}^2\mathrm d\mu \leqslant 1$$ hence $f$ is also an element of the consider set.