= Solution
Here $F=\mathbf1_A$ is bounded and hence lies in $L^2$ as well as $L^1$. Conditional expectation on a sub-<sigma-algebra> is the <orthogonal projection> in $L^2$ onto its measurable subspace. We can use these projections to justify both limits without any symmetry assumption on the <probability measure>.
The increasing coordinate sigma-algebras $\mathcal E_n$ generate the whole <product sigma-algebra> $\mathcal F$. Functions depending on finitely many coordinates are dense in $L^2(\mathcal F)$: their cylinder indicators generate $\mathcal F$, and the usual monotone-class approximation proves density of their linear span. If a cylinder function $V$ is $\mathcal E_N$-measurable, then for $n\geq N$ the $L^2$ projection contraction gives
$$
\|F-\mathbb E[F\mid\mathcal E_n]\|_2
\leq\|F-V\|_2+\|\mathbb E[V-F\mid\mathcal E_n]\|_2
\leq2\|F-V\|_2.
$$
Approximate $F$ arbitrarily well by such $V$ to prove the first $L^2$ limit.
For the decreasing tail sigma-algebras $\mathcal G_n$, put $Y_n=\mathbb E[F\mid\mathcal G_n]$. If $m\geq n$, the tower property gives $\mathbb E[Y_n\mid\mathcal G_m]=Y_m$. Projection orthogonality therefore yields
$$
\|Y_n-Y_m\|_2^2=\|Y_n\|_2^2-\|Y_m\|_2^2.
$$
The norms on the right decrease and are nonnegative, so $(Y_n)$ is Cauchy in $L^2$ and has a limit $Y$. Choose an almost-surely convergent subsequence and take its pointwise upper limit. For every fixed $n$ its tail consists of $\mathcal G_n$-measurable functions, so this supplies a version of $Y$ measurable with respect to every $\mathcal G_n$, hence to $\mathcal G=\bigcap_n\mathcal G_n$. For any $C\in\mathcal G$, $\mathbb E[Y_n\mathbf1_C]=\mathbb E[F\mathbf1_C]$; passing to the limit identifies $Y=\mathbb E[F\mid\mathcal G]$.
Finally $\|W\|_1\leq\|W\|_2$ on a <probability space>, by the <Cauchy-Schwarz inequality>. Thus
$$
\boxed{\mathbb E[F\mid\mathcal E_n]\longrightarrow F,\qquad
\mathbb E[F\mid\mathcal G_n]\longrightarrow\mathbb E[F\mid\mathcal G]\quad\text{in }L^1.}
$$
These are the increasing-filtration and <reverse martingale convergence theorem> limits; the argument above establishes the requested norm convergence directly.
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