# Applying a formula like $\mathbb{E}[ X \vert \sigma(\mathcal{F},\mathcal{G}) ] = \mathbb{E}[X \vert \mathcal{G}]$

Let $(Y_1, \ldots, Y_n)$ be a $[0,1]^n$-valued random vector and $U_1, \ldots, U_n$ independent random variables, uniformly distributed on $[0,1]$, and independent of $(Y_1, \ldots, Y_n)$. For some fixed $j \in \{1, \ldots, n\}$ consider the random variable $X := \mathbb{1}_{\{U_j \leq Y_j\}}$. Then I want to show that $$\mathbb{E}[X \vert \sigma(Y_1, \ldots, Y_n)] = \mathbb{E}[X \vert \sigma(Y_j)].$$ To prove this, I thought I can use the result, that if $\mathcal{F}$ is independent of $\sigma(\mathcal{G},\sigma(X))$, then $$\mathbb{E}[ X \vert \sigma(\mathcal{F},\mathcal{G}) ] = \mathbb{E}[X \vert \mathcal{G}].$$ Hence we set $\mathcal{G} : = \sigma(Y_j)$ and $\mathcal{F} : = \sigma(Y_i \colon i \in \{1, \ldots, n\} \setminus \{j\} )$. Then $\sigma(\mathcal{F},\mathcal{G}) = \sigma(Y_1, \ldots, Y_n)$. Now it remains to show that $\mathcal{F}$ is independent of $$\sigma(\mathcal{G},\sigma(X)) = \sigma(Y_j,\mathbb{1}_{\{ U_j \leq Y_j \}}).$$ But I only know that $U_j$ is independent of $\mathcal{F}$. I don't know anything about $Y_j$ and $Y_i$ for $i \neq j$. How can I safe this?

If $V$ and $W$ are two independent vectors (say of dimension $m$ and $k$) and $f\colon\mathbb R^m\times \mathbb R^k\to\mathbb R$ is measurable, then $$\mathbb E\left[f\left(V,W\right)\mid \sigma\left(W\right) \right]=g\left(W\right)$$ where $g$ is defined by $$g\left(w_1,\dots,w_k\right)=\mathbb E\left[f\left(V,w_1,\dots,w_k\right) \right].$$ Apply this to the following cases:
1. $m=1$, $k=n$, $f\left(v,w_1,\dots,w_n \right) =\mathbf 1\left\{v\leqslant w_j \right\}$, $V=U_j$ and $W=\left(Y_1,\dots,Y_n\right)$.
2. $m=k=1$, $f\left(v,w \right) =\mathbf 1\left\{v\leqslant w\right\}$, $V=U_j$ and $W= Y_j$.