I imagine this is standard, but I can't manage to prove it. Any help/hints much appreciated!
Let $B$ be a real, unital, separable Banach algebra and $(X,F,\mu)$ a finite measure space. Suppose that $f,g, f\cdot g:X\to B$ are Bochner integrable, where $f\cdot g$ denotes the pointwise product. Denote by $\mu_f,\mu_g\in\mathrm{Meas}(B)$ the pushforward measures $\mu_f(U)=\mu(f^{-1}(U))$ and similarly $\mu_{(f,g)}\in\mathrm{Meas}(B\times B)$, $\mu_{(f,g)}(U\times V)=\mu\{x\in X:(f(x),g(x))\in U\times V\}.$
Suppose that $$\mu_{(f,g)}=\mu_f\otimes\mu_g$$ (so $f$ and $g$ independent if $\mu$ were a probability measure). Do we then have that
$$\int_Xfd\mu\int_Xgd\mu=\int_Xfgd\mu\;?$$
In the case $B=\mathbb R$ and say wlog $f,g\geq 0$, we can approximate with simple functions $f_n=\sum_0^{n^2}\frac{k}{n}\chi_{f^{-1}[k/n,(k+1)/n)}$ and $g_n$ similarly. It is then easy to show that $\int f_ng_n=\int f_n\int g_n$ and fiddling with MCT & DCT completes the proof. I'm having trouble extending this type of reasoning to general Banach algebras.
The same argument works. Suppose first that $f$ and $g$ are simple, i.e. $$f=\sum_{i=1}^n\chi_{A_i}a_i,\quad g=\sum_{j=1}^m\chi_{B_j}b_j,$$ and so $$f\cdot g=\sum_{i=1}^n\sum_{j=1}^m\chi_{A_i\cap B_j}a_i\cdot b_j$$ which is also simple. Taking Bochner integrals we find $$\int_Xf\cdot gd\mu=\sum_{i=1}^n\sum_{j=1}^m\mu(A_i\cap B_j)a_i\cdot b_j,\\ \left(\int_Xfd\mu\right)\cdot\left(\int_Xgd\mu\right)=\left(\sum_{i=1}^n\mu(A_i)a_i\right)\cdot\left(\sum_{j=1}^m\mu(B_j)b_j\right)$$ and so clearly the result will follow if we have $\mu(A_i\cap B_j)=\mu(A_i)\mu(B_j)$ for all $i,j$. But $A_i=f^{-1}\{a_i\}$ and $B_j=g^{-1}\{b_j\}$, and so $$\mu(A_i\cap B_j)=\mu_{(f,g)}(\{a_i\}\times\{b_j\})=\mu_f\{a_i\}\mu_g\{b_j\}=\mu(A_i)\mu(B_j).$$
For the general case, recall that for separable spaces measurable functions are strongly measurably. Since $f$ is $\sigma(f)$-measurable, $f$ is $\sigma(f)$-strongly measurable, thus Bochner integrable with respect to $(X,\sigma(f),\mu|_{\sigma(f)})$. Hence there exists a sequence $\{f_n\}$ of $\sigma(f)$-measurable simple functions such that $\int_X\|f_n-f\|d\mu\to0)$. By passing to a subsequence, we may assume $\|f_n-f\|\to0$ almost everywhere, and by multiplying by $\chi_{\{x\in X:\|f_n(x)\|\le2\|f(x)\|\}}$ we may assume $\|f_n\|\le2\|f\|$. Define $\{g_n\}$ similarly.
Observe that $\mu_{(f_n,g_n)}=\mu_{f_n}\otimes\mu_{g_n}$ for each $n$, $f_n\cdot g_n\to f\cdot g$ almost everywhere (by continuity of multiplication), and $\|f_n\cdot g_n\|\le\|f_n\|\|g_n\|\le4\|f\|\|g\|$. We want to be able to conclude \begin{align*} \int_Xf\cdot gd\mu &=\lim_{n\to\infty}\int_Xf_n\cdot g_nd\mu\\ &=\lim_{n\to\infty}\left(\int_Xf_nd\mu\right)\cdot\left(\int_Xg_nd\mu\right)\\ &=\left(\int_Xfd\mu\right)\cdot\left(\int_Xgd\mu\right). \end{align*} The only line we need to justify is the first, and this will follow from the dominated convergence theorem if we can show that $\|f\|\|g\|$ is integrable. However, we also have that $\mu_{(\|f\|,\|g\|)}=\mu_{\|f\|}\otimes\mu_{\|g\|}$, and so by the real case we have $$\int_X\|f\|\|g\|d\mu=\left(\int_X\|f\|d\mu)\right)\left(\int_X\|g\|d\mu\right)<\infty,$$ completing the proof.