Let $f:X\to B$ be a function defined on a measurable space $(X,\Sigma)$ and taking values in a Banach space $B$. The function is called $\Sigma$-simple if it is of the form $f=\sum_{n=1}^N x_n 1_{A_n}$ with $A_n\in \Sigma$ for each $1\leq n\leq N$. The function is called strongly $\Sigma$-measurable if it is the pointwise limit of a sequence of $\Sigma$-simple functions.
A corollary of the Pettis mesurability theorem is that the pointwise limit of a sequence of strongly $\Sigma$ -measurable functions is strongly $\Sigma$-measurable.The notes am reading use the preceding result to prove the following:
Proposition. If $f:X\to B$ strongly-$\Sigma$ measurable and $\phi:B\to F$ is continuous, where $F$ is another Banach space, then $\phi\circ f$ is strongly-$\Sigma$ measurable.
Proof. Choose $\Sigma$-simple functions $f_n$ converging to $f$ pointwise. Then $\phi\circ f_n\to \phi\circ f$ pointwise and the result follows from the previous corrollary.
I don't see why we need the force of the Pettis theorem to prove this. If $f_n$ is $\Sigma$-simple, then we can choose a representation $f_n=\sum_{k=1}^N x_k 1_{A_k}$ where the $A_k\in \Sigma$ are disjoints and cover $X$. Then $\phi\circ f_n=\sum_{k=1}^N \phi(x_k) 1_{A_k}$ is itself a $\Sigma$-simple function. From $\phi\circ f_n\to \phi\circ f$ and the definition of strong measurability we get the result.
Am I missing something?
Best Answer
Yes, you are right. If $f_n$ are $\Sigma$-simple functions, then $\phi\circ f_n$ are $\Sigma$-simple functions. Since $\phi$ is continuous, if $f_n$ converges to $f$ a.e. then $\phi\circ f_n$ converges to $\phi \circ f$. So, $\phi \circ f$ is strong measurable.