Let $I_1,\ldots,I_m$ be a partition of $I:=[0,1)$ such that $c:=\lambda(I_1)=\cdots=\lambda(I_m)$, where $\lambda$ is the Lebesgue measure on $\mathcal B(\mathbb R)$. Let $f:[0,1)\to\mathbb R$ be Lebesgue integrable and $$\mathcal U_D=\frac{1_D}{\lambda(D)}\lambda$$ denote the uniform distribution on any $D\in\mathcal B(\mathbb R)$ with $\lambda(D)>0$.
Let $U$ and $U_j$ be real-valued random variables distributed according to $\mathcal U_I$ and $\mathcal U_{I_j}$, respectively. How are $\operatorname{Var}[f(U)\mid U\in I_j]$ and $\operatorname{Var}[f(U_j)]$ related?
What I obtain is \begin{equation}\begin{split}\operatorname{Var}[f(U)\mid U\in I_j]&=\operatorname E\left[\left(f(U)-\operatorname E\left[f(U)\mid U\in I_j\right)\right)^2\mid U\in I_j\right]\\&=\operatorname E\left[f(U_j)^2\right]-2\operatorname E[f(U_j)]+\operatorname E\left[f(U)\mid U\in I_j\right]^2\operatorname P[U\in I_j]\end{split}\tag1\end{equation} while $$\operatorname{Var}[f(U_j)]=\operatorname E[f(U_j)^2]-\operatorname E[f(U_j)]^2\tag2.$$ Am I calculating something wrong?
Let $I$ be an interval on $\mathbb{R}^+$ and $I_j\in \mathscr{B}(I)$. Then if $U\sim \textrm{Uniform}$ on $I$ and $U_j\sim \textrm{Uniform}$ on $I_j$ then $$E[g(U)|U \in I_j]=\frac{E[g(U)\mathbf{1}_{\{U \in I_j\}}]}{P(U \in I_j)}=\frac{1}{\lambda(I_j)\lambda(I)}\int_{I_j}g(x)dx=\frac{1}{\lambda(I)}E[g(U_j)]$$ If $I=[0,1)$ then it follows that: $$V[f(U)|U\in I_j]\stackrel{(*)}{=}\frac{1}{\lambda(I)}E[f^2(U_j)]-\frac{1}{\lambda(I)^2}E[f(U_j)]^2=V[f(U_j)]$$
$(*)$: recall $$\begin{aligned}V[X|A]&=E[(X-E[X|A])^2|A]=\\ &=E[X^2|A]+(E[X|A])^2-2E[X|A]^2=\\ &=E[X^2|A]-E[X|A]^2\end{aligned}$$