How to prove that $P(X_n = x | \bigcup_{i = 1}^g (X_k = j_i) \cap \bigcup_{i = 1}^m A_i) = P(X_n = x | \bigcup_{i = 1}^g (X_k = j_i))$ in MarkovChain? $n > k > k_1 > \dots > k_r > 0$ are natural numbers, let $A_1, A_2, \dots, A_m$ be events such that $A_i = (X_{k_1} = a_{i1}, \dots X_{k_r} = a_{ir})$, where X_n is Markov chain and $\not\exists i, j: A_i = A_j$. $j_i$ are different numbers.
I already proved that $P(X_n = x | X_k = j \cap \bigcup_{i = 1}^m A_i) = P(X_n = x | X_k = j)$. I managed to prove $P(X_n = x | \bigcup_{i = 1}^g (X_k = j_i) \cap \bigcup_{i = 1}^m A_i) = \sum_{i = 1}^g (P(X_n = x | X_k = j_i)\cdot P(X_k = j_i, \bigcup_{i = l}^m A_l))\frac{1}{P(\bigcup_{i = 1}^g X_k = j_i \cap \bigcup_{l = 1}^m A_l)}$
but can't go any further.
Main idea I was using is $P(A|B) = P(B|A)\frac{P(A)}{P(B)}$
 A: This is just an idea (not yet an answer though) with $g=2$. It should be straight-forward to generalize though.
Let $A,B_1,B_2$ be events and consider the addition law of probability with the definition of the conditional probability
$$P\left( B_1\cup B_2 | A \right)P\left( A \right)=P\left( (B_1 \cup B_2) \cap A \right)=P\left( (B_1\cap A)\cup(B_2\cap A) \right) \\
=P\left( B_1\cap A \right)+P\left( B_2\cap A \right)-P\left( (B_1\cap B_2) \cap A \right) \\
=P\left( B_1|A \right)P(A)+P\left( B_2|A \right)P(A)-P\left( B_1\cap B_2|A \right)P(A)$$
and canceling $P(A)\neq 0$, we arrive at $$P\left(B_1 \cup B_2|A\right)=P\left( B_1|A \right)+P\left( B_2|A \right)-P\left( B_1\cap B_2|A \right) \, .$$
Now consider Bayes theorem
$$P\left( (B_1\cup B_2)\cap A \right)=P\left( A|B_1\cup B_2 \right)P(B_1\cup B_2)
=P\left( B_1\cup B_2|A \right)P(A)\\
=\left\{ P\left( B_1|A \right) + P\left( B_2|A \right) - P\left( B_1\cap B_2|A \right) \right\}P(A) \\
=P\left( A | B_1 \right)P(B_1) + P\left( A | B_2 \right)P(B_2) - P\left( A | B_1\cap B_2 \right)P(B_1\cap B_2)$$
and so
$$P\left( A|B_1\cup B_2 \right)=\frac{P(B_1)}{P(B_1\cup B_2)} \, P(A|B_1) + \frac{P(B_2)}{P(B_1\cup B_2)} \, P(A|B_2) \\ - \frac{P(B_1 \cap B_2)}{P(B_1\cup B_2)} \, P(A|B_1\cap B_2)\,. \tag{1}$$
If you set $$A\equiv (X_n=x) \\ B_1\equiv(X_k=j_1)\cap \bigcup_{i=1}^m A_i \\ B_2\equiv(X_k=j_2)\cap \bigcup_{i=1}^m A_i$$
you can use your result $P(A|B_1)=P(A|X_k=j_1)$ already obtained (similarly for $B_2$). Since $X_k=j_1$ and $X_k=j_2$ can not be true at the same time (since $j_1$ and $j_2$ are distinct), the last term in (1) with $P(B_1\cap B_2)$ vanishes. It remains an idea for my above comment.
