Is the following set closed in Lp? $\mathsf{L}^{p}(\Omega,\mathcal{F},\mathrm{P})$ is the
collection of random variables such that $\operatorname{E}(\left\vert
X\right\vert ^{p})<\infty$.
Let $\mathsf{A}$ be a set of all random variable $X$ such that
$\operatorname{E}(X)=1$, $X\geq0$, and $X\in\mathsf{L}^{p}(\Omega
,\mathcal{F},\mathrm{P})$, $p>1$.
Let $\mathsf{B}$ be linear subspace span by a set of $N$ random variables,
such that $\mathsf{A}\cap\mathsf{B}=\emptyset$. Is it $\mathsf{A}-\mathsf{B}$
a closed set in $\mathsf{L}^{p}(\Omega,\mathcal{F},\mathrm{P})$? Where $\mathsf{A}-\mathsf{B}\equiv\{a-b:a\in\mathsf{A},b\in\mathsf{B}\}$.
 A: Yes, this set is closed. In fact, we do not need all of your assumptions,
for example the assumption $A\cap B=\emptyset$ is superflous.
Let $\left(X_{n}-Y_{n}\right)_{n\in\mathbb{N}}$ be a sequence with
$X_{n}\in A$, $Y_{n}\in B$ and $X_{n}-Y_{n}\xrightarrow[n\to\infty]{}Z$,
where convergence is in $L^{p}$. We have to show $Z\in A-B$. To
this end, observe that $L^{p}\left(P\right)\hookrightarrow L^{1}\left(P\right)$
and furthermore that (by definition of $A$), we have
$$
\left\Vert X_{n}\right\Vert _{L^{1}}\;\overset{X_{n}\geq0}{=}\;\mathbb{E}\left(X_{n}\right)=1
$$
for all $n\in\mathbb{N}$. Furthermore, since convergent sequences
are bounded, we have $\left\Vert X_{n}-Y_{n}\right\Vert _{L^{1}}\leq C$
for some $C>0$ and all $n\in\mathbb{N}$.
By the triangle inequality, this implies 
$$
\left\Vert Y_{n}\right\Vert _{L^{1}}\leq\left\Vert Y_{n}-X_{n}\right\Vert _{L^{1}}+\left\Vert X_{n}\right\Vert _{L^{1}}\leq1+C
$$
for all $n\in\mathbb{N}$. Thus, the sequence $\left(Y_{n}\right)_{n\in\mathbb{N}}$
is a bounded sequence in the finite-dimensional(!) space $B$ (equipped
with the norm $\left\Vert \cdot\right\Vert _{L^{1}}$) and thus admits
a convergent subsequence, i.e. $\left\Vert Y_{n_{k}}-Y\right\Vert _{L^{1}}\xrightarrow[k\to\infty]{}0$
for some $Y\in B\subset L^{p}$. Since on the finite-dimensional(!)
space $B$, all norms are equivalent, we even get $\left\Vert Y_{n_{k}}-Y\right\Vert _{L^{p}}\xrightarrow[k\to\infty]{}0$.
Now, we conclude
$$
\left\Vert X_{n_{k}}-\left(Z+Y\right)\right\Vert _{L^{p}}\leq\left\Vert \left(X_{n_{k}}-Y_{n_{k}}\right)-Z\right\Vert _{L^{p}}+\left\Vert Y_{n_{k}}-Y\right\Vert _{L^{p}}\xrightarrow[k\to\infty]{}0,
$$
so that $X_{n_{k}}\xrightarrow[k\to\infty]{}Z+Y$. Now, since $L^{p}\hookrightarrow L^{1}$,
it is an easy exercise to show that the set $A$ is closed in $L^{p}$
(to show that nonnegativity is preserved, use that an $L^{p}$-convergent
sequence admits an almost everywhere convergent subsequence). Hence,
$Z+Y\in A$ and thus
$$
Z=\left(Z+Y\right)-Y\in A-B,
$$
so that $A-B\subset L^{p}$ is closed.
