Suppose $f$ is a measurable function. Let B be any measurable set. I was reading a text on ergodic theory and it states that: $\chi_{f^{-1}(B)} =\chi_B \circ f$. I don't understand why this is true. Why is the composition of $\chi_B$ and $f$ equal to the left-hand side? Thanks.
2026-05-16 02:36:41.1778899001
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$$(\chi_B\circ f)(a)=\chi_B(f(a)) =\cases{1&if $f(a)\in B$\\ 0&otherwise} =\cases{1&if $a\in f^{-1}(B)$\\ 0&otherwise}=\chi_{f^{-1}(B)}(a).$$