Exercise 1.3.5 (Equivalent formulation of a simple function)

Question

Show that an unsigned function $f:\mathbb{R}^d	o[0,+\infty]$ is a simple function if and only if it is measurable and takes on at most finitely values.

Elu Thingol · Accepted Answer

\begin{itemize}\item[$\implies$)] $f$ is measurable, since it is trivially the limit of the constant sequence equal to itself; and it takes on finitely many values by definition.
\item[$\impliedby$)] Now suppose that $f$ is measurable with the finite image $f(\mathbb{R}^d) = \{c_1,\ldots,c_n\}$. To show that $f$ is measurable it suffices to show that $f^{-1}(\{c_i\})$ is measurable for an arbitrary $c_i$. But $\{c_i\}$ is closed in $\mathbb{R}$, so its inverse image $\{c_i\}$ under a measurable function $f$ must be measurable by Lemma 1.3.9 (xi).
\end{itemize}

Exercise 1.3.5 (Equivalent formulation of a simple function)

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Exercise 1.3.5 (Equivalent formulation of a simple function)

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