# Are continuous functions strongly measurable?

Measure theory is still quite new to me, and I'm a bit confused about the following.

Suppose we have a continuous function $f: I \rightarrow X$, where $I \subset \mathbb{R}$ is a closed interval and $X$ is a Banach space. I can show that $f$ is weakly measurable: for each $v \in X^*$, we have that the mapping $x \mapsto v(f(x))$ is continuous since it is a composition of continuous mappings $v$ and $f$ (is this correct?).

I also know that a continuous function is measurable. In this case, does measurable mean the same as strongly measurable? If it does, how can you show this using Pettis' theorem (i.e. why is there a null set $N\subset I$ such that the set $\{f(x) | x\in I\backslash N\}$ is separable)? Or is it easier to prove it without Pettis' theorem?

For continuous $f$, does strong convergence then also imply that it is summable / Bochner integrable, since the mapping $x \mapsto ||f(x)||$ is the composition of continuous maps ($f$ and $||.||$)? EDIT: This last sentence is nonsense of course, $x \mapsto ||f(x)||$ must be summable, not continuous.

• If you are new to measure theory, then you should take a beginning course in measure theory, not a course of measure theory in Banach space. – TCL Apr 27 '13 at 18:31
• Do the case $I = [0,1]$ first. Use that $f(I)$ is a compact metric space. – Martin Apr 27 '13 at 19:02
• Is this correct: $I=[0,1]$ is compact, hence $f(I)$ is a compact subset of a metric space and therefore separable. By Pettis' theorem it follows that any continuous function is strongly measurable? – ScroogeMcDuck Apr 28 '13 at 11:18
• Correction: any continuous function with compact domain is strongly measurable? – ScroogeMcDuck Apr 28 '13 at 11:24
• Yes, this is correct. The general case follows from this since a closed interval is a countable union of compact intervals. // However, you do not need to use Pettis's theorem. Note that $f$ is uniformly continuous on $[0,1]$. Use this to approximate $f$ by step functions. // Your proof of weak measurability is okay, btw. // Please use @Martin in your future comments so I am notified. – Martin Apr 29 '13 at 16:40

Let $$0=t_0\le t_1\le...\le t_{n}=1$$ and suppose that $$\Delta=\max_{i\in\{1,...,n\}}(t_{i}-t_{i-1})\to 0$$ as $$n\to \infty$$. Let $$E_{i}=[t_{i-1}, t_i)$$ and for each $$i=1,...,n$$ fix $$\xi_{i}\in E_i$$. Let us now define $$f_{n}(t)=\sum_{i=1}^{n}f(\xi_{i})\chi_{E_{i}}(t)$$ for all $$t\in [0,1]$$ and $$n\in \mathbb{N}$$. Fix $$t\in [0,1]$$, $$\varepsilon>0$$ and observe that $$\|f(t)-f_{n}(t)\| \le \sum_{i=1}^{n}\|f(t)-f(\xi_{i})\|\chi_{E_{i}}(t)=\|f(t)-f(\xi_{i(t)})\|<\varepsilon$$ for sufficiently large $$n$$ which is a consequance of uniform continuity of $$f$$ on compact set $$[0,1]$$. This imples that $$\lim_{n\to\infty}\|f(t)-f_{n}(t)\|=0$$ for all $$t\in [0,1]$$.