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I would greatly appreciate it if you kindly give me some feedback on this question.

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    Is $f^n$ the function that takes $x$ to $f(f(f(\cdots f(x)\cdots)))$ or to $f(x)^n$? – 2017-02-22
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    And: Do you mean for some $n>4$ or for all $n>4$? – 2017-02-22
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    Here $f(x)^n$ like $f^2$. – 2017-02-22
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    $f=(f^5)^{1/5}$ and $x \mapsto x^{1/5}$ is measurable. – 2017-02-22
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    Why $f^2$ is measurable we can not say $f$ is measurable then? – 2017-02-22

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As Shalop said, since $f^5$ is measurable, we have \begin{align*} \{f \leq a\} = \{f^5 \leq a^5\} \end{align*} which is measurable for any $a \in \mathbb{R}$.

To address your comment, it is not clear how to write $\{f \leq a\}$ as an expression of sets of the form $\{f^2 \leq b\}$ with $b \in \mathbb{R}$, since $\{f^2 \leq b\} = \{|f| \leq b^{1/2}\}$, not $\{f \leq b^{1/2}\}$.

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    So could we say for $f^{2k+1}$ it is true? – 2017-02-23