Is there an isometry between the direct sum of $L^1(\mu_i)$ and $L^1(\nu)$, i.e. $\bigoplus_{i} L^1(\mu_i)\cong L^1(\nu)$?

Question

In https://math.stackexchange.com/a/74877/653080 it is mentioned that $\mathcal{M}(K)\cong L^1(\nu)$ for some measure $\nu$, whereas $\mathcal M(K)$ is an $\mathcal{l}_1-$sum of $L_1(\mu)$ spaces is it mentioned that $\mathcal{M}(K)\cong \bigoplus_{i\in I} L^1(\mu_i)$ for some mutually singular probability measures $\mu_i$. This hints at that $\bigoplus_{i\in I} L^1(\mu_i)\cong L^1(\nu)$.

Does this mean that in general there exists a (probability) measure $\nu$ such that $\bigoplus_{i\in I} L^1(\mu_i)\cong L^1(\nu)$, when $\mu_i$ are mutually singular probability measures $\mu_i$? If so, can we choose $\nu$ to be any probability measure $\nu$ with full support on $K$?

The spaces $\bigoplus_{i\in I} L^1(\mu_i)$ and $L^1(\nu)$ are defined as $$ L^1(\nu) = \{f:K\to\mathbb{R}^d \;|\; \int_K|f(x)|d\nu(x) < \infty \} \\ \bigoplus_{i\in I} L^1(\mu_i) = \{ (f_1,f_2...) \;|\; f_i\in L^1(\mu_i)\} $$ with norms $$ ||f||_{L^1(\nu)} = \int_K|f(x)|d\nu(x) \\ ||f||_{\bigoplus_{i\in I} L^1(\mu_i)} = \sum_{i\in I}||f_i||_{L^1(\mu_i)} $$

Answer 1

No, this does not work. In particular, note that the space $(K, \mathcal{B}(K), \nu )$ is a finite (hence, separable) measure space. This implies (e.g. see Brezis Theorem 4.13) that $L^1(\nu )$ is a separable Banach space, but then it cannot be isometrically isomorphic to $\mathcal{M}(K)$, which is non-separable in general.