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.
Best Answer
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 $t\in [0,1]$.