Let $t>0$ and $E$ a Lebesgue measurable set with $\mu(E)=t$. If $s\in(0,t)$, show that it exists $A\subset E$ with $\mu(A)=s$.
I can show the claim if $E$ is contained in a finite closed interval [-a,a] which is in general not the case.
Is it true that $E$ is contained in such an interval almost everywhere, i.e. that $\mu(E\cap [-a,a])=t$ and $\mu(E\setminus [-a,a])=0$? I think then my proof would still work.
On
The Theorem that you're looking to use is the continuity of measure. If $\{E_n\}$ is a sequence of ascending measurable sets, then
$$\lim_{n\to\infty}\mu(E_n)=\mu\left(\bigcup_{n=1}^\infty E_n\right)$$
Similarly, if $\{F_n\}$ is a sequence of descending measurable sets AND one of the sets $F_n$ has finite measure, then
$$\lim_{n\to\infty}\mu(F_n)=\mu\left(\bigcap_{n=1}^\infty F_n\right)$$
The goal is to construct a sequence of sets $E_1\subseteq E_2\subseteq E_3\subseteq...$ where the measures of the sets approach $s$. In doing so, we will also construct a sequence of sets $F_1\supseteq F_2\supseteq F_3\supseteq ...$ where the measure approaches $s$.
First, we consider the sets $G_k$ for $k=0,1,2,...$ where
$$G_k=E\cap[-k,k]$$
Since these sets are ascending and union to $E$, we must have
$$\lim_{k\to\infty}\mu(G_k)=\mu(E)=t$$
As such, for large enough $k$, $\mu(G_k)>s$. This means that we can pick some $k$ such that $\mu(G_k)\leq s$ and $\mu(G_{k+1})>s$. Take $E_0=G_k$ and $F_0=G_{k+1}$.
For convenience, we will also examine a sequence of real numbers $a_n$, where we define $a_0=0$.
For the next sets of the sequences, we consider the set
$$E\cap[-k-1/2,k+1/2]$$
If this set has measure no greater than $s$, we define this set as $E_1$, $F_1=F_0$, and $a_1=1/2$. Otherwise, define this set as $F_1$, $E_1=E_0$, and $a_1=a_0$. The number $a_1$ satisfies
$$E_1=E\cap[-k-a_1,k+a_1]\;\;\;\;\;\;\;\;F_1=E\cap[-k-a_1-1/2,k+a_1+1/2]$$
We continue this process inductively. We consider the set
$$E\cap[-k-a_{n-1}-1/2^n,k+a_{n-1}+1/2^n]$$
Depending on the measure of this set, we either assign this to $E_n$ or $F_n$ while leaving the other one the same. We also update $a_n=a_{n-1}+1/2^n$ if we assign the set to $E_n$ or assign $a_n=a_{n-1}$ otherwise. This means that at every step of the way,
$$E_n=E\cap[-k-a_n,k+a_n]\;\;\;\;\;\;\;\;F_n=E\cap[-k-a_n-1/2^n,k+a_n+1/2^n]$$
and
$$\mu(E_n)\leq s<\mu(F_n)$$
I'm going to skip some details here that you can hopefully reason out, but this formulation leads to
$$\lim_{n\to\infty}\mu(E_n)=\lim_{n\to\infty}\mu(F_n)$$
As such, the limit must be $s$. By the continuity of measure, the union of all $E_n$ (or the intersection of all $F_n$ must have measure $s$).
This is a bit of a long winded explanation, but it is a constructive proof, which I find to be (sometimes) easy to digest. Hopefully this helps!
On
Liapounoff's theorem says that the range of an atom-free vector measure is a closed convex set.
Here's Halmos' proof:
https://projecteuclid.org/download/pdf_1/euclid.bams/1183511896
Answering the explicit question in your post: no, $E$ is not always contained in such an interval almost everywhere. Consider $$ E = \bigcup_{n = 1}^\infty \left[n, n + \frac{1}{2^n}\right]. $$ Then $E$ is Lebesgue-measurable, and $\mu(E) = \sum_{n = 1}^\infty \frac{1}{2^n} = 1$, but for every finite closed interval $[-a, a]$, there is an $n > a$ with $[n, n + 2^{-n}] \cap [-a, a] = \emptyset$. Consequently, $$ \mu(E \setminus [-a, a]) \geq \mu([n, n + 2^{-n}]) = 2^{-n} > 0. $$
EDIT: I also thought about the result you want to prove and found the following general result: given a measurable set $E$, the map $$ \mu_E: \mathbb{R}^+ \to \mathbb{R}^+: x \mapsto \mu(E \cap [-x, x]) $$ is uniformly continuous. Indeed, let $x < y \in \mathbb{R}^+$. Then, first of all, $|\mu_E(y) - \mu_E(x)| = \mu_E(y) - \mu_E(x)$, by monotonicity of $\mu$. Also, \begin{align*} \mu_E(y) - \mu_E(x) &= \mu(E\cap[-y, y]) - \mu(E \cap [-x, x]) \\ &= \mu( (E \cap [-x, x]) \cup (E \cap ([-y, -x) \cup (x, y]))) - \mu(E \cap [-x, x]) \\ &= \mu(E \cap [-x, x]) + \mu(E \cap ([-y, -x) \cup (x, y])) - \mu(E \cap [-x, x]) \\ &= \mu(E \cap ([-y, -x) \cup (x, y])), \end{align*} where the penultimate equality follows from the fact that $E \cap [-x, x]$ and $E \cap ([-y, -x) \cup (x, y])$ are disjoint sets. Now, as $E \cap ([-y, -x) \cup (x, y]) \subseteq [-y, -x) \cup (x, y]$, we have $$ \mu(E \cap ([-y, -x) \cup (x, y]) \leq \mu([-y, -x) \cup (x, y]) = 2(y - x). $$ Thus, we find that $|\mu_E(y) - \mu_E(x)| \leq 2(y - x)$, and by symmetry, we get $$ |\mu_E(y) - \mu_E(x)| \leq 2 |y - x| $$ for all $x, y \in \mathbb{R}^+$. It follows that $\mu_E$ is uniformly continuous.
How does the desired result follow from this? Note that $\mu_E(0) = 0$ and that $\lim_{x \to +\infty} \mu_E(x) = \mu(E)$. As $\mu_E$ is continuous, it follows from the intermediate value theorem that for each $s \in [0, \mu(E))$, there is an $x \in \mathbb{R}^+$ such that $\mu_E(x) = s$.