For each positive integer $n$, let $f(n)$ denote the smallest possible value of $$|A_1 \cup A_2 \cup \dots \cup A_n|$$ where $A_1, A_2, A_3 \dots A_n$ are sets such that $A_i \not\subseteq A_j$ and $|A_i| \neq |A_j|$ whenever $i \neq j$. Determine $f(n)$ for each positive integer $n$.
To determine \( f(n) \) for each positive integer \( n \), we need to find the smallest possible value of \( |A_1 \cup A_2 \cup \dots \cup A_n| \) where \( A_1, A_2, \ldots, A_n \) are sets such that \( A_i \not\subseteq A_j \) and \( |A_i| \neq |A_j| \) whenever \( i \neq j \).

1. **Base Cases:**
   - For \( n = 1 \):
     \[
     f(1) = 0
     \]
     This is because if we have only one set, it can be the empty set, and the union of an empty set is also empty.
   
   - For \( n = 2 \):
     Suppose \( A_1, A_2 \subseteq [1, 2] \). The sets must have distinct cardinalities, so one set must have cardinality 1 and the other must have cardinality 2. However, this is not possible since the set with cardinality 2 would include the set with cardinality 1, violating the condition \( A_i \not\subseteq A_j \). Therefore, we need at least 3 elements to construct such sets. For example, \( A_1 = \{1\} \) and \( A_2 = \{2, 3\} \):
     \[
     f(2) = 3
     \]

   - For \( n = 3 \):
     Suppose \( A_1, A_2, A_3 \subseteq [1, 4] \). The sets must have distinct cardinalities 1, 2, and 3. However, if one set has cardinality 3, it would include the set with cardinality 2, violating the condition. Therefore, we need at least 5 elements to construct such sets. For example, \( A_1 = \{1, 4\} \), \( A_2 = \{2, 3, 4\} \), and \( A_3 = \{5\} \):
     \[
     f(3) = 5
     \]

2. **Inductive Step:**
   - **Lemma 1:** Suppose \( A_1, \ldots, A_n \subseteq [1, n+2] \) are good with cardinalities \( 1, \ldots, n \). Then we can construct the same for \( n+2 \). Define:
     \[
     B_k := A_k \cup \{n+3\}, \quad B_{n+1} := [1, n+2], \quad B_{n+2} := \{n+4\}
     \]
     This construction ensures that \( B_1, \ldots, B_{n+2} \) are good sets with distinct cardinalities \( 1, \ldots, n+2 \). Therefore, \( f(n+2) \le n+4 \).

   - **Lemma 2:** Suppose \( n \ge 3 \) and \( f(n) \le n+1 \). Then also \( f(n-1) \le n \). Suppose \( A_1, \ldots, A_n \subseteq [1, n+1] \) is good. The sets \( A_1, \ldots, A_{n-1} \) must have distinct cardinalities \( 1, \ldots, n-1 \). Without loss of generality, let \( A_n = \{n+1\} \). Then \( A_1, \ldots, A_{n-1} \subseteq [1, n] \) are good sets, implying \( f(n-1) \le n \).

3. **Conclusion:**
   Suppose \( f(n) \le n+1 \) for some \( n \ge 3 \). Then by Lemma 2, \( f(n-1) \le n \), and continuing this process, we get \( f(3) \le 4 \), which contradicts our earlier result that \( f(3) = 5 \). Therefore, \( f(n) = n+2 \) for \( n \ge 3 \).

The final answer is \( \boxed{ f(n) = n+2 } \) for \( n \ge 3 \).