Uniqueness of The Law of Iterated Expectations

Question

Let $x$ be a random variable with distribution $f(x)$ on a bounded set $P\in\mathbb{R}$ with mean $\mathbb{E}[x]=\mu$. Let $P_j$, for $j\in\{1,2,\dots,J\}$, be partitions of $P$ such that $\cup_j P_j=P$ and $P_j \cap P_k = \emptyset$. Define \begin{align*} m(P_j)=\int \mathbb{1}\{x \in P_j\}f(x) dx \qquad \mu(P_j)=\int x \,\,\mathbb{1}\{x \in P_j\}\frac{f(x)}{m(P_j)} dx \end{align*} Where $m(P_j)$ represents the mass of partition $P_j$, while $\mu(P_j)$ the conditional expectation. Naturally $\sum_{j=1}^J m(P_j)=1$. Furthermore, by the law of iterated expectations we have \begin{align*} \sum_{j=1}^J m(P_j) \mu(P_j) = \mu \end{align*} That is, to retrieve the unconditional mean, the "only" way to combine these conditional means, is to weight them by the measure of the sets themselves. Consider alternative "weights" \begin{align*} \sum_{j=1}^J f_j\left(m(P_1),m(P_2),\dots,m(P_J)\right) = 1 \end{align*} That is $f_j:[0,1]^K \to [0,1]$. Furthermore, we know that $f_j(...)\neq m(P_j)$ for at least one $j$. Hence, surely these new "weights" differ from the original ones. Define the new convex combination of the conditional expectations as follows \begin{align*} \sum_{j=1}^J f_j\left(m(P_1),m(P_2),\dots,m(P_J)\right) \mu(P_j) \end{align*} I want to show that this new expression will "generically" not be equal to $\mu$, or if it is, these will be "rare cases".