Is $f^{-1} (\alpha) = \bigcap_{m=0}^\infty \bigcup_{n=m+1}^\infty f_n^{-1} (\alpha)$ correct?

Question

Let $(f_n)$ be a sequence of functions from $X$ to $\mathbb R$. We define $f:X \to \mathbb R$ by $f(x) = \limsup_{n \to \infty} f_n (x)$. Then we write $f = \limsup_{n \to \infty} f_n$. Assume that $f(x)$ is finite for all $x \in X$.

In this comment, @TheBridge said that $$f^{-1} (\alpha) = \bigcap_{m=0}^\infty \bigcup_{n=m+1}^\infty f_n^{-1} (\alpha)$$

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I've tried but unable to come up with a formula with intersection and union as @TheBridge did. IMHO, that formula doesn't seem right to me.

Could you please confirm if @TheBridge's formula is correct or not? Thank you so much!

Answer 1

False. Take $f_n(x)=\frac 1 n $ for all $n$, $f(x)=0$ and $\alpha =0$.