If $P$ has marginals $P_1, P_2$, is $L^1(P_1) + L^1(P_2)$ closed in $L^1(P)$? Suppose that  $\mathbb{X}=\mathbb{X}_1\times \mathbb{X}_2$ and suppose that $ P$ is a probability measure on  $\mathbb{X}$ with marginals $ P_i$ on $\mathbb{X}_i, i=1,2$, i.e., $$\int f_i(x_i)\, dP=\int f_i(x_i)\, dP_i \,\,\,\,\text{for  }i=1,2,$$ for $P_i$ intergrable functions $f_i(x_i)$, $i=1,2$ (looking at them as functions of $(x_1,x_2)$). Is the linear space spanned by $L^1(P_1)\cup L^1(P_2)$ ($L^1(P_1) + L^1(P_2)$) closed in $L^1(P)$?
 A: No, it need not be closed. I'll give an example with $L^1(P_1)+L^1(P_2)$ dense in $L^1(P)$ but not equal to $L^1(P)$.
Take $\mathbb{X}_1=\mathbb{X}_2=\mathbb{N}=\{0,1,2,\ldots\}$. Next, choose sequences $\epsilon_n,q_n > 0$ with $\sum_{n=0}^\infty q_n=1$, $\epsilon_n < 1/2$ and $\epsilon_n\to0$ as $n\to\infty$.
Define the measure $P$ on $\mathbb{X}=\mathbb{N}^2$ by $P(\{(i,j)\})=p_{i,j}$ where
\begin{align}
&p_{2n,2n} = q_n\epsilon_n,\\
&p_{2n,2n+1}=q_n\left(\frac12-\epsilon_n\right),\\
&p_{2n+1,2n}=q_n\frac12,
\end{align}
and all other $p_{m,n}$ are zero. These sum to $1$. If $f\colon\mathbb{N}^2\to\mathbb{R}$ then we can write $f(m,n)=f_1(m)+f_2(n)$ on the support of $P$ where,
\begin{align}
&f_1(2n)=0,\\
&f_1(2n+1)=f(2n+1,2n)-f(2n,2n),\\
&f_2(2n)=f(2n,2n),\\
&f_2(2n+1)=f(2n,2n+1).
\end{align}
If $f$ has finite support then so do $f_1$ and $f_2$, so they are in $L^1(P_1)$ and $L^1(P_2)$ respectively, and $f\in L^1(P_1)+L^1(P_2)$. It follows that for $f\in L(P)$ setting $g_N(m,n)=1_{\{m,n\le N\}}f(m,n)$ gives $g_N\in L^1(P_1)+L^1(P_2)$ and $g_N\to f$ in $L^1(P)$. So, $L^1(P_1)+L^1(P_2)$ is dense in $L^1(P)$.
Next, define $f\in L^1(P)$ by $f(2n,2n)=\epsilon_n^{-1}$ and all other $f(m,n)$ equal to zero. Then, $\lVert f\rVert_1=\sum_n q_n=1$, so $f\in L^1(P)$. However, if $f(m,n)=f_1(m)+f_2(n)$ on the support of $P$, then
$$
\epsilon_n^{-1}=f(2n,2n)-f(2n+1,2n)=f_1(2n)-f_1(2n+1).
$$
So, $\lvert f_1(2n)\rvert+\lvert f_1(2n+1)\rvert\ge\epsilon_n^{-1}$. So,
\begin{align}
\lVert f_1\rVert &= \sum_{n=0}^\infty\left(P_1(2n)\lvert f_1(2n)\rvert+P_1(2n+1)\lvert f_1(2n+1)\rvert\right)\\
&=\sum_{n=0}^\infty\frac{q_n}{2}\left(\lvert f_1(2n)\rvert+\lvert f_1(2n+1)\rvert\right)\\
&\ge\frac12\sum_{n=0}^\infty p_n\epsilon_n^{-1}.
\end{align}
If $\sum_n p_n\epsilon_n^{-1}=\infty$ then $f_1\not\in L^1(P_1)$. In particular, we can take $p_n=2^{-n}$ and $\epsilon_n=2^{-n-2}$, showing that $L^1(P_1)+L^1(P_2)\not=L(P)$.
