Given the following exercise:
Let $n\in \mathbb N$. For $t\in \mathbb R$ and $M\subset \mathbb R^n$ define $$M_t = \{x\in \mathbb R^{n-1} : (x,t) \in M\}.$$ Now let $L, K\in \mathcal L^n$. Proof that if $\forall t\in \mathbb R: \lambda_{n-1}(L_t) = \lambda_{n-1}(K_t)$, then $\lambda_n(L) = \lambda_n(K)$.
Where $\mathcal L^n$ denotes the set of Lebesgue-measurable sets in $\mathbb R^n$ and $\lambda_n$ is the n-dimensional Lebesgue measure.
Now I guess a possible proof looks like
$$\lambda_n(L) = \int_\mathbb R \lambda_{n-1}(L_t)d\lambda_1 = \int_\mathbb R \lambda_{n-1}(K_t)d\lambda_1 = \lambda_n(K)$$
but I guess this is not enough (even for me it is not quite clear how these steps were made). Could someone help adding some steps to make this more rigorous?