Exercise about conditional probabilities 
Exercise: Let $S_1$ and $S_2$ be two disjoint events, and let $\mathcal{G}$ be a sub-$\sigma$-algebra. Show that the following hold with probability $1$.
  \begin{align*}
  \mathrm{(a)}& \quad 0 \le \mathbf{P}[S_1 | \mathcal{G}] \le 1 \\
  \mathrm{(b)}& \quad \mathbf{P}[S_1 \uplus S_2 | \mathcal{G}] = \mathbf{P}[S_1 | \mathcal{G}] + \mathbf{P}[S_2 | \mathcal{G}]
\end{align*}

Solution: (a) First we show that $0 \le \mathbf{P}[S_1 | \mathcal{G}] $.
Define 
\begin{equation*}
  G:= \bigl\{ \omega \in \Omega : \mathbf{P}[S_1 | \mathcal{G}](\omega) < 0\bigr\} \, ,
\end{equation*}
and assume that $\mathbf{P}[G] > 0$. 
We see that 
\begin{equation*}
  \bigl\{\mathbf{P}[S_1|\mathcal{G}] < - 1/n\bigr\} \uparrow G\, .
\end{equation*}
Since $\mathbf{P}$ is a measure it must be lower semicontinuous, so 
\begin{equation*}
  \mathbf{P}\bigl[\mathbf{P}[S_1|\mathcal{G}] < - 1/n\bigr] \overset{n\rightarrow\infty}{\longrightarrow} \mathbf{P}\bigl[G\bigr] > 0\, ,
\end{equation*}
which means that for a certain $N \in \mathbb{N}$: 
\begin{equation*}
  H:= \bigl\{\mathbf{P}[S_1|\mathcal{G}] < - 1/N\bigr\} \Longrightarrow \mathbf{P}\bigl[H\bigr] > 0 \, .
\end{equation*}
Now we get
\begin{equation*}
  \mathbf{E}\bigl[\mathbf{P}[S_1 | \mathcal{G}]\, \mathbf{1}_{H}\bigr] < - \mathbf{P}[H] / N < 0 \, ,
\end{equation*}
but from the definition of $\mathbf{P}[S_1 | \mathcal{G}]\, $ it should follow that
\begin{equation}
  \mathbf{E}\bigl[\mathbf{P}[S_1 | \mathcal{G}]\, \mathbf{1}_{H}\bigr] = \mathbf{P}[S_1 \cap H] \geq 0 \, ,
\end{equation}
which is a contradiction.
The proof that $\mathbf{P}[S_1 | \mathcal{G}] \leq 1$ is very similar, I've done it, but I'll omit it for now.
(b) This follows, because we have almost surely \begin{align*}
  \mathbf{E}\bigl[\mathbf{P}[S_1 \uplus S_2 | \mathcal{G}]\, \mathbf{1}_{G}\bigr] &= \mathbf{P}[(S_1 \uplus S_2)\cap G] = \mathbf{P}\bigl[(S_1 \cap G) \uplus (S_2 \cap G)\bigr] = \\
  &= \mathbf{P}[S_1 \cap G] + \mathbf{P}[S_2 \cap G] = \mathbf{E}\bigl[\mathbf{P}[S_1 | \mathcal{G}] \, \mathbf{1}_{G} \bigr] +  \mathbf{E}\bigl[\mathbf{P}[S_2 | \mathcal{G}]\, \mathbf{1}_{G} \bigr] \, .
\end{align*}

Could you please check if my proof is correct? 

Thank you very much!
 A: For (a) -- I would just comment that $H \in \mathcal{G}$ Because,
$$\int_{\Omega} \mathbb{E}[\mathbf{1}_{[S_1]} | \mathcal{G}] \mathbf{1}_{[H]} dP(\omega) = \int_{\Omega} \mathbf{1}_{[S_1]} \mathbf{1}_{[H]} dP(\omega)$$
If and only if $H \in \mathcal{G}$. It is clear it is by measurability and properties of $\sigma-$algebra. Additionally,  since $P$ is a measure it has continuity of measure so for any monotonic increasing or decreasing sequence of sets $A_n \uparrow A$ or $A_n \downarrow A$, it is that $\lim\limits_{n \rightarrow \infty} P(A_n) = P(\lim\limits_{n \rightarrow \infty} A_n) = P(A)$. 
For (b) -- I would prove it like this: 
Let $X$ and $Y$ be random variables defined on $(\Omega, \mathcal{F}, P)$ and $\mathcal{G} \subset \mathcal{F}$. Note that $\mathbb{E}[X | \mathcal{G}] + \mathbb{E}[Y | \mathcal{G}]$ is $\mathcal{G}$ measurable. Then, for all $A \in \mathcal{G}$, 
$\begin{align} 
\int_A \big(\mathbb{E}[X | \mathcal{G}] + \mathbb{E}[Y | \mathcal{G}]\big) dP(\omega) &= \int_A \mathbb{E}[X | \mathcal{G}] dP(\omega) + \int_A \mathbb{E}[Y | \mathcal{G}] dP(\omega) & \text{Linearity of Integration}  \\
& = \int_A X dP(\omega) + \int_A Y dP(\omega) & \text{Def. Cond. Expect.} \\
& = \int_A \big(X + Y\big) dP(\omega) & \text{Linearity of Integration} \\
& = \int_A \mathbb{E}[X + Y | \mathcal{G}] dP(\omega) & \text{Def. Cond. Expec.} 
\end{align} $
This shows $\mathbb{E}[X + Y | \mathcal{G}] = \mathbb{E}[X | \mathcal{G}] + \mathbb{E}[Y | \mathcal{G}]$ almost surely. 
For your problem, take $X = \mathbf{1}_{[S_1]}$ and $Y = \mathbf{1}_{[S_2]}$. Then, 
$$P(S_1 \cup S_2 | \mathcal{G}) := \mathbb{E}[\mathbf{1}_{[S_1 \cup S_2]} | \mathcal{G}] = \mathbb{E}[\mathbf{1}_{[S_1]} + \mathbf{1}_{[S_2]} | \mathcal{G}]$$
And the result holds almost surely. 
A: If you define conditional probabilités using conditional expectations, as $\mathbf{P}(A | \mathcal{G}) = \mathbf{E}[\mathbf{1}_A |\mathcal{G}]$, things look simpler : 
By the monotony property of expectations, as $0 \leq \mathbf{1}_A \leq 1$, we have
$$0 \leq \mathbf{E}[\mathbf{1}_S | \mathcal{G}] \leq 1$$
By the linearity, if $A$ and $B$ are disjoints, then $\mathbf{1}_{A \cup B} = \mathbf{1}_A + \mathbf{1}_B$, therefore
$$\mathbf{E}[\mathbf{1}_{A \cup B} |\mathcal{G}] = \mathbf{E}[ \mathbf{1}_A|\mathcal{G}] =  + \mathbf{E}[\mathbf{1}_B |\mathcal{G}] $$ 
All the following inequalities/equalities are only true almost everywhere.
But you're maybe supposed to prove these properties (monotonicity) ?
