Right-continuity of completed filtration I have a question about filtration.
Now fix a measurable space $(\Omega,\mathcal{M})$. Let $(\mathcal{M}_{t})_{t\in[0,\infty)}$ be a filtration on $(\Omega,\mathcal{M})$.
We set 
\begin{eqnarray*}
\mathcal{M_{t+}}=\cap_{s>t} \mathcal{M}_{s}
\end{eqnarray*}
I want to show that
\begin{eqnarray*}
(\mathcal{M_{t+}})^{P}=(\mathcal{M}^{P}_{t})_{+}
\end{eqnarray*}
Here $P$ is a probability measure on $(\Omega, \mathcal{M})$. $(\mathcal{M_{t+}})^{P},\mathcal{M}^{P}_{t}$ are completion of  $\mathcal{M_{t+}},\mathcal{M}_{t}$ w.r.t. $P$ respectively.
i.e. 
$(\mathcal{M}_{t+})^{P}=\{A \subset \Omega | A \Delta B \subset N{\rm\,for\,some\,} B \in \mathcal{M_{t+}}, N\in\mathcal{M},P(N)=0   \}$
$\mathcal{M}_{t}^{P}=\{A \subset \Omega | A \Delta B \subset N{\rm\,for\,some\,} B \in \mathcal{M_{t}}, N\in\mathcal{M},P(N)=0   \}$
$\cdot$ Proof of $(\mathcal{M_{t+}})^{P} \subset(\mathcal{M}^{P}_{t})_{+}$
Let $A \in(\mathcal{M_{t+}})^{P}  $.  By definition, there exists $B \in \mathcal{M_{t+}}, N\in\mathcal{M},P(N)=0$ such that $A \Delta B\subset N$. Since $B \in \mathcal{M_{s}} $ for all $s>t$, $A\in \mathcal{M}_{s}^{P}$ for all $s>t$. This implies $A\in(\mathcal{M}^{P}_{t})_{+}$.
$\cdot$ Proof of $(\mathcal{M}^{P}_{t})_{+} \subset (\mathcal{M_{t+}})^{P} $ (Unfinished)
Let $A \in (\mathcal{M}^{P}_{t})_{+}(=\cap_{s>t}\mathcal{M}_{s}^{P})$. Then for all $s>t$, we can find $B_{s} \in  \mathcal{M}_{s}, N_{s} \in \mathcal{M},P(N_{s})=0$  such that $A \Delta B_{s} \subset N_{s}$. 
$p,q>t$:fix
If $B_{p}\Delta B_{q}$ is negligible,then $B_{p} \in (\mathcal{M}_{t}^{P})_{+}$.
Can I deduce $B_{p} \in (\mathcal{M}_{t+}$? 
I don't know what to do...
 A: Denote by $$\mathcal{N} := \{N \subseteq \Omega; \exists M \in \mathcal{M}: N \subseteq M, \mathbb{P}(M)=0\}$$ the subsets of $\mathbb{P}$-null sets.

Let $A \in (\mathcal{M}_{t}^{\mathbb{P}})_+$. Then there exists for any $k \in \mathbb{N}$ some $B_k \in \mathcal{M}_{t+1/k}$ such that $$N_k := A \Delta B_k \in \mathcal{N}$$ We set $$B := \bigcup_{n \in \mathbb{N}} \bigcap_{k \geq n} B_k.$$
Since $$\bigcap_{k \geq n} B_k \in \mathcal{M}_{t+1/n}$$ and $\bigcap_{k \geq n} B_k$ is increasing (in $n$), we have
$$B = \bigcup_{n \in \mathbb{N}} \bigcap_{k \geq n} B_k = \bigcup_{n \geq N} \bigcap_{k \geq n} B_k \in \mathcal{M}_{t+1/N}$$
for each $N \in \mathbb{N}$. Hence, $B \in \mathcal{M}_{t+}$. Moreover,
$$\begin{align*} A \setminus B &= A \cap B^c = A \cap \bigcap_{n \in \mathbb{N}} \bigcup_{k \geq n} B_k^c = \bigcap_{n \in \mathbb{N}} \bigcup_{k \geq n} \underbrace{(B_k^c \cap A)}_{A \setminus B_k \subseteq N_k}. \end{align*}$$
Since the union is countable, we find that $A \setminus B \in \mathcal{N}$. A similar argument shows that $B \setminus A \in \mathcal{N}$. Consequently, $A \Delta B \in \mathcal{N}$. Hence, $A \in (\mathcal{M}_{t+})^{\mathbb{P}}$. This finishes the proof.
