Below we use Bochner measurability and Bochner integral. Let
- $(X, \mathcal A, \mu)$ and $(Y, \mathcal B, \nu)$ be complete $\sigma$-finite measure spaces,
- $(E, | \cdot |)$ a Banach space,
- $S (X)$ the space of $\mu$-simple functions from $X$ to $E$,
- $L^0 (X)$ the space of $\mu$-measurable functions from $X$ to $E$,
- $L^1 (X)$ the space of $\mu$-integrable functions from $X$ to $E$,
- $Z := X \times Y$,
- $\mathcal C$ the product $\sigma$-algebra of $\mathcal A$ and $\mathcal B$,
- $\lambda$ the product measure of $\mu$ and $\nu$,
- $(Z, \overline{\mathcal C}, \overline{\lambda})$ the completion of $(Z, \mathcal C, \lambda)$.
For $\delta >0$ and $f,g \in L^0 (Y)$, we write $$ \begin{align*} \{|f - g| > \delta\} &:= \{y \in Y : |f (y) - g(y)| > \delta\}, \\ \nu (|f - g| > \delta) &:= \nu (\{|f - g| > \delta\}). \end{align*} $$
For $f, g \in L^0 (Y)$, we define $$ \hat \rho_Y (f, g) := \inf_{\delta >0} \{ \nu (|f - g| > \delta) +\delta \}. $$
Then $\hat \rho_Y$ is an extended pseudometric on $L^0 (Y)$. For $f_n, f \in L^0(Y)$, we have $\hat \rho_Y (f_n, f) \to 0$ IFF $f_n \to f$ in measure.
My understanding It seems to me $(L^0 (Y), \hat \rho_Y)$ is not necessarily a metric linear space because scalar multiplication is not necessarily continuous in the topology induced by $\hat \rho_Y$. Take $E=Y = \mathbb R$ and $\nu$ its Lebesgue measure. Let $f_n := 1$ and $\lambda_n = 1-\frac{1}{n}$. Then $|\lambda_n -1| \to 0$ and $\hat \rho_Y(f_n, 1) =0$. However, $$ \hat \rho_Y (\lambda_n f_n, 1) = \hat \rho_Y \bigg (\frac{1}{n}, 0 \bigg) = +\infty. $$
Could you confirm if my above understanding is fine?
