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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
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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

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