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We say that $f$ is Borel measurable if for $c \in \mathbb{R}$, $\{ x \in E : f(x) > c \}$ is a Borel set.

Let $B$ be a Borel measurable set, and let $f$ be a continuous strictly increasing function, then $f^{-1}(B)$ is a Borel set.

My goal is to first show that the given set is a $\sigma$-algebra, then I need to show that it is in fact Borel measurable.

To show it is a $\sigma$-algebra, this comes from the fact that continuous functions preserve the following:

  1. $f^{-1}(E_n^c) = (f^{-1}(E_n))^c $.

  2. $f^{-1}(E_j \setminus E_k) = f^{-1}(E_j) \setminus f^{-1}(E_k)$.

  3. $f^{-1}(\bigcup E_i) = \bigcup f^{-1}(E_i)$.

I think the last step is just to show that $f^{-1}((a, \infty))$ is Borel measurable.

I'm in a bit of need of inspiration as to how to complete this proof.

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Your title does not reflect what you do in the question. Note that you haven't really used continuity, yet. Points 1,2 and 3 hold for all functions, continuous or not. For your last step note that $f^{-1}((a,\infty))$ is open, hence Borel, by continuity of $f$. However, you haven't used the assumption that $f$ be strictly increasing. This makes me suspect that your actual task is to show the statement in the title. For this, just note that a strictly increasing continuous $f$ has a continuous inverse $g$. Then apply what you did so far to $g$. Done. – t.b. Sep 17 '11 at 18:31
Sorry, I goofed slightly: that continuous inverse $g$ exists only on the image of $f$ which is an open interval of the form $(a,b)$, $(a,\infty)$ or $(-\infty,b)$ but that doesn't make it much harder. – t.b. Sep 17 '11 at 18:43

1 Answer 1


As t.b. said, for this part you don't need $f$ to be strictly increasing: the statement holds for all continuous functions. See, for example, Prove that any continuous function is Borel Measurable

Strictly increasing continuous functions map Borel sets to Borel sets

To handle this question (stated in the title), consider the inverse function $g=f^{-1}$. It is a continuous function, defined on the range of $f$ (an interval of $\mathbb R$). Apply the first part to $g$.

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