# Bochner integral of a function from $[0,1]$ to $c_0$

I want to compute the Bochner integral of the function $$f: [0,1]\to c_0\,, \quad f(x)= \left( \frac{\cos nx}{n} \right)_{n\in\mathbb N}.$$

First, I need to check for strong measurability. Since $$c_0$$ is separable, strong and weak measurability coincide. However, how do I check whether the preimages of Borel sets (or open sets) are Borel? It seems very hard to think about the preimages of sets of null sequences.

For now, I'll just assume that it is measurable. Then $$f$$ is Bochner integrable if the function $$x \mapsto \|f(x)\|_\infty = \sup_{n \in \mathbb N} \left\vert\frac{\cos nx}{n}\right\vert$$ is integrable. This is, of course, true because $$\cos$$ is bounded and thus $$\cos(nx)/n \to 0$$.

Now, how can I compute $$\int_{[0,1]} f\,\mathrm d\lambda$$? By definition, I would need to find a sequence of simple functions $$f_n: [0,1] \to c_0$$ that converges to $$f$$ or the Hahn-Banach theorem could be used?

The dual of $$c_0$$is $$\ell^{1}$$. To check weak measurability of $$f$$ we have to check that $$\sum a_n \frac {\cos nx} n$$ is measurable for each $$(a_n) \in \ell^{1}$$. This is true because the partial sums are continuous functions, the sum is (uniformly!) convergent and limits of Borel measurable functions are Borel measurable.
Bochner integral (BI) coincides with Pettis integral. So the BI of $$f$$ is an element $$(b_n)$$ of $$c_0$$ such that $$\sum \int_0^{1} a_n \frac {\cos nx} ndx =\sum a_nb_n$$ for each $$(a_n) \in \ell^{1}$$. Obviously this gives $$b_n=\int_0^{1} \frac {\cos nx} ndx=\frac {\sin n} {n^{2}}$$