# Equality of two Lebesgue Volumes

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?

• Shouldn't that be $\mathbb R^{n-1}$ in the definition of $M_t?$ – zhw. Apr 7 '17 at 16:43
• Of course, I fixed it. – Staki42 Apr 7 '17 at 16:46

You can use Fubini's Theorem to $\chi_L$, the characteristic function of $L$, but it seems like overkill for your question. It is enough to show that \begin{align} \lambda_n (L)=\int _{\mathbb{R}} \lambda_{n-1} (L_t) d\lambda_{1} \end{align}
We can easily show this for a rectangle, and it is extended to an open set $V$; if $V$ is a countable union of pairwise non-overlapping rectangles $\{ A_i \}$,
\begin{align} \int _{\mathbb{R}} \lambda_{n-1} (V_t) d\lambda_{1} &=\int _{\mathbb{R}}\sum _i \lambda_{n-1} (V_t \cap A_i) d\lambda_{1}\\ &=\sum _i \int _{\mathbb{R}}\lambda_{n-1} (V_t \cap A_i) d\lambda_{1}\\ &=\sum _i \lambda_{n-1} (A_i)=\lambda_n (V) \end{align} (We can change integral and sum, since every function on $t$ here is simple function.)
Now assume that $L$ is a bounded measurable set. Then there are countable open sets $\{ V_k\}$ such that $V_k \downarrow L$. Notice that $V_{k,t} \downarrow L_t$, thus \begin{align} \int _{\mathbb{R}} \lambda_{n-1} (L_t) d\lambda_{1} &=\int _{\mathbb{R}} \lim_{k\to\infty}\lambda_{n-1} (V_{k,t}) d\lambda_{1}\\ &=\lim_{k\to\infty}\int _{\mathbb{R}} \lambda_{n-1} (V_{k,t}) d\lambda_{1}\\ &=\lim_{k\to\infty} \lambda_{n}(V_{k})= \lambda_{n}(L)\\ \end{align} (Again, we can change limit and integral, since $\lambda_{n-1} (V_{k,t})$ is a sum of simple functions.)
Finally, for general case, consider $L \cap B(\mathbf{0},n)$ and take $n\to\infty$.