For any $t \in \mathbb{R}$ let $K_t$ be a measurable subset of $\mathbb{R}^n$ with $0$ Lebesgue measure. Let also $K := \bigcup_{t \in \mathbb{R}}{\{t\} \times K_t}$
I know that if $K$ is measurable then $K$ has measure $0$, this comes from the Fubini's Theorem since
$$\mathcal{L}^{n+1}(K) = \int_{\mathbb{R}^{n+1}}{\chi_{K} d\mathcal{L}^{n+1} } = \int_{\mathbb{R}}{ \int_{\mathbb{R}^n}{ \chi_K(t,x) d\mathcal{L}^{n}(x)} d\mathcal{L}^1(t)} = \int_{\mathbb{R}}{ \mathcal{L}^{n}(K_t) d\mathcal{L}^1(t) } = \int_{\mathbb{R}}{ 0 \mathcal{L}^1(t) } = 0$$
My question is if $K$ is always measurable.
This would make sense since the Lebesgue measure space is complete, and in particular when the outer measure of a set is $0$ then the set is Lebesgue measurable, therefore a way to prove this could be to prove, without using Fubini's theorem, that the outer measure of $K$ is $0$.