For a given Lebesgue measurable set $E$ and a continuous function $f\colon E \longrightarrow \overline{\mathbb{R}},\ $ I am asked to prove that $f$ is Lebesgue measurable.
My Proof: If $f$ is Lebesgue measurable, the set $A = \{ x \in E \mid f(x)<\alpha \}$ is Lebesgue measurable for every $\alpha \in \mathbb{R}$. (H. L. Royden, Real Analysis)
Then, I need to show that $A = f^{-1} ((-\infty,\ \alpha))$ is Lebesgue measurable for every $\alpha\in \mathbb{R}$.
I begin with $(-\infty,\ \alpha)$, which is an open set of $\overline{\mathbb{R}}$ for every $\alpha \in \mathbb{R}$.
Then, by continuity of $f,\ $ $f^{-1} ((-\infty,\ \alpha))$ is open in $E\ $ for every $\alpha \in \mathbb{R}$.
Open sets are Lebesgue measurable, hence $A = f^{-1} ((-\infty,\ \alpha))$ is Lebesgue measurable. $\qquad \qquad$ (Q.E.D.)
Is there any problem with the proof?