Open Set Criterion for Lebesgue Measurable Functions

Open Set Criterion for Lebesgue Measurable Functions

Recall from the Lebesgue Measurable Functions page that an extended real-valued function $f$ with Lebesgue measurable domain $D(f)$ is said to be a Lebesgue measurable function if for all $\alpha \in \mathbb{R}$, the following sets are Lebesgue measurable sets:

(1)
\begin{align} \quad \{ x \in D(f) : f(x) < \alpha \} \end{align}

We will now look at simple theorem which gives us criterion for a function to be Lebesgue measurable.

Theorem 1: Let $f$ be an extended real-valued function defined with Lebesgue measurable domain $D(f)$. Then $f$ is a Lebesgue measurable function if and only if for every open set $O$, the inverse image $f^{-1}(O)$ is a Lebesgue measurable set.
  • Proof: Let $f$ be an extended real-valued function with Lebesgue measurable domain $D(f)$.
  • $\Rightarrow$ Suppose that $f$ is a Lebesgue measurable function. Then for all $\alpha \in \mathbb{R}$ the following sets are Lebesgue measurable:
(2)
\begin{align} \quad \{ x \in D(f) : f(x) < \alpha \} \end{align}
  • Let $O$ be an open set. Then $O$ can be expressed as a countable union of bounded open intervals. That is, $\displaystyle{O = \bigcup_{n=1}^{\infty} I_n}$ where for each $n \in \mathbb{N}$, $I_n = (a_n, b_n)$ where $-\infty < a_n < b_n < \infty$. For each $I_n$ let $B_n = (-\infty, b_n)$ and let $A_n = (a_n, \infty)$. Then $I_n = A_n \cap B_n$. So:
(3)
\begin{align} \quad f^{-1}(O) & = f^{-1} \left ( \bigcup_{n=1}^{\infty} I_n \right ) \\ & = \bigcup_{n=1}^{\infty} f^{-1}(I_n) \\ & = \bigcup_{n=1}^{\infty} f^{-1}((a_n, b_n)) \\ &= \bigcup_{n=1}^{\infty} f^{-1} (A_n \cap B_n) \\ &= \bigcup_{n=1}^{\infty} [f^{-1} (A_n) \cap f^{-1}(B_n)] \\ &= \bigcup_{n=1}^{\infty} [\{ x \in D(f) : f(x) > a_n \} \cap \{ x \in D(f) : f(x) < b_n \}] \end{align}
  • So $f^{-1}(O)$ is a countable union of Lebesgue measurable sets so $f^{-1}(O)$ is a Lebesgue measurable set.
  • $\Leftarrow$ Suppose that for every open set $O$, $f^{-1}(O)$ is a Lebesgue measurable set. Then open interval $(-\infty, \alpha)$ is an open set for all $\alpha \in \mathbb{R}$. Therefore, for all $\alpha \in \mathbb{R}$ we have that the following sets are Lebesgue measurable:
(4)
\begin{align} \quad f^{-1}((-\infty, \alpha)) = \{ x \in D(f) : f(x) < \alpha \} \end{align}
  • So $f$ is a Lebesgue measurable function. $\blacksquare$
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