My question refers to functions with values in Banach spaces and under what conditions the limit of a sequence of measurable functions is also measurable.

But first, let me recall some well-known results from real analysis. Let $(X,\mathcal{F})$ be a measurable space, $\bar{\mathbb{R}}$ the extended real line, and $\mathcal{B}(\bar{\mathbb{R}})$ the standard Borel $\sigma$-algebra generated by the open sets of $\bar{\mathbb{R}}$.

(1) If a sequence of measurable functions $f_n:(X,\mathcal{F})\rightarrow(\bar{\mathbb{R}},\mathcal{B}(\bar{\mathbb{R}}))$ converges to a function $f$ pointwise in $X$, then $f$ is also measurable w.r.t. $\mathcal{F}$ and $\mathcal{B}(\bar{\mathbb{R}})$. (See, e.g., Rudin, Real and Complex Analysis, Theorem 1.14.)

(2) This can be generalized to the case of a complete measure space $(X,\mathcal{F},\mu)$ and convergence of $f_n$ to $f$ pointwise-$\mu$-a.e. in $X$. (This is stated without proof in Hunter and Nachtergaele, Applied Analysis, Theorem 12.24. Can anyone point to a proof of this, or is it trivial?)

Dudley (Real Analysis and Probability, Theorem 4.2.2) generalizes part (1) above to functions with values in metric spaces. Therefore (1) also holds for functions with values in a Banach space $(Y,\Vert\cdot\Vert)$, with corresponding Borel $\sigma$-algebra $\mathcal{B}(Y)$. My question is: can part (2) be also generalized? In other words, does the following result hold:

Let $(X,\mathcal{F},\mu)$ be a complete measure space and $(Y,\Vert\cdot\Vert)$ be a Banach space. Let $f_n:(X,\mathcal{F})\rightarrow(Y,\mathcal{B}(Y))$ be a sequence of measurable functions that converges to a function $f$ pointwise-$\mu$-a.e. in $X$ (i.e., $\Vert f_n(x) - f(x)\Vert \rightarrow 0$ as $n\rightarrow\infty$, for $x$ $\mu$-a.e. in $X$), then $f$ is also measurable w.r.t. $\mathcal{F}$ and $\mathcal{B}(Y)$.

Note, I use "measurable" in the standard sense: if $U\in\mathcal{B}(Y)$ then $f^{-1}(U)\in\mathcal{F}$.

It seems to me that the result should hold but I haven't been able to find a proof in the standard references. Is it trivial? Also, why should the measure space $(X,\mathcal{F},\mu)$ be complete? What fails if it is not complete. Finally, if the result holds, does this mean that standard measurability is equivalent to so-called strong or Bochner measurability? (Maybe this last question should be the topic of another post.)

Thanks in advance!


Yes, it's rather trivial. You might note the following fact: if $f$ is measurable, $\mu$ is complete, and $f = g$ $\mu$-a.e., then $g$ is measurable. This is valid for functions from any complete measure space to any other measurable space. (Proving it is a good exercise.) Essentially, when you have a complete measure, you can change a function arbitrarily on a null set without breaking its measurability.

Now suppose $f_n \to f$ $\mu$-a.e., with each $f_n$ measurable. Let $A$ be the set of $x$ such that $f_n(x) \to f(x)$. Note $A$ is a measurable set since $A^C$ is $\mu$-null and hence measurable (by completeness of $\mu$). Let $$g_n = \begin{cases}f_n(x), & x \in A \\ 0, & x \notin A\end{cases}$$ Then clearly $g_n$ is measurable. Now $g_n(x)$ converges for all $x$, so the limit $g(x) = \lim_{n \to \infty} g_n(x)$ is measurable, as you know. But $g(x) = f(x)$ on $A$, so $f = g$ a.e. By the fact above, $f$ is measurable.

If your underlying measure space is not complete, certainly things can go wrong. For a trivial example, let $B$ be any set which is $\mu$-null but not measurable. If $f_n$ is the zero function for all $n$, we have $f_n \to 1_B$ a.e., but $1_B$ is not a measurable function.

  • $\begingroup$ Thanks so much for your response. Regarding the statement: if $f$ is measurable, $\mu$ is complete, and $f=g$ $\mu$-a.e., then $g$ is measurable. The proof is in Royden (Real Analysis, 2nd. ed., pg. 68). $\endgroup$ Jan 20 '11 at 2:12
  • $\begingroup$ A different proof to the one given by Royden is as follows: $\endgroup$ Jan 20 '11 at 2:14
  • $\begingroup$ Let $A=\{x\in X| f(x)=g(x)\}$ so that $\mu(A^c)=0$ and let $B\in \mathcal{B}(Y)$. Then $g^{-1}(B)=(g^{-1}(B)\cap A)\cup(g^{-1}(B)\cap A^c)=(f^{-1}(B)\cap A)\cup(g^{-1}(B)\cap A^c)$. Since $g^{-1}(B)\cap A^c$ is a subset of $A^c$ it must be measurable because $\mu$ is complete. The other sets $A$ and $f^{-1}(B)$ are also measurable. Therefore, $g^{-1}(B)$ must be measurable. $\endgroup$ Jan 20 '11 at 2:22

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