Infimum inequality comparing restrictions.

Question

Suppose $f$ is a continuous function on the real line. Say we have two collections of sets $\{A_k\}_{k=1}^{n}$ and $\{B_k\}_{k=1}^{m}$, where $n>m$ and \begin{align} \bigcap_{k=1}^{n} A_k &= \bigcap_{k=1}^{m} B_k = \emptyset \\ \bigcup_{k=1}^{n}A_k &= \bigcup_{k=1}^{m}B_k \end{align} Can I say that $$ \frac{1}{n}\sum_{k=1}^{n} \inf_{x\in A_k}f(x) \ge \frac{1}{m}\sum_{k=1}^{m} \inf_{x\in B_k}f(x). $$ Intuitively the answer is yes, since the $B_k$ give less restrictions, thus the average infimum should be lower.

Answer 1

Take $f$ the identity function. Let $m=2$ with $B_1 = [0,1]$ and $[1,2]$. Let $n = 3$ with $A_1= [0,1], A_2 = [0,1]$ and $A_3 = [1,2]$. Now your inequality reads $$\frac{1}{3} \geq \frac{1}{2}$$ and we have a contradiction.

EDIT: This is a counterexample to the original question. For a counterexample to the modified one see the comments.