Prove that the set of reachable values is convex. Let $f:[0,1] \to \mathbb R^n$ be an integrable function.
Prove that the set $$\left\{\int_A f(x)\ \mathrm dx: A\subset [0,1] \text { is measurable}\right\}$$ is convex.
Proving the statement ends up being equivalent to proving that $$\forall \alpha\in[0,1]: \exists A_\alpha\ \text{s.t } \int_{A_\alpha} f = \alpha \int_0^1 f$$
My first idea was to try and find a set $A_\alpha$ with density $\alpha$ on every subset of $[0,1]$, but that turned out to be impossible for non trivial cases.
So I tried to approach $f$ with piecewise constant functions on $[0,1]$, but I couldn't find a way to tell anything relevant about $f$'s set of reachable values from the sets of the functions that approach it.
 A: Let $\mathcal F\subseteq L^1([0,1])^*$ be the set of a.e. equivalence classes of functions $F:[0,1]\to [0,1],$ equipped with the weak-* topology. $\mathcal F$ is a compact set by the Banach-Alaoglu theorem, and it's convex.
The image of $\mathcal F$ under the map $f^*:L^1([0,1])^*\to\mathbb R^n$ defined by $F\mapsto \int fF$ is a compact convex set. This image obviously contains the set defined in the question. We want to show that any point $y\in f^*(\mathcal F)$ can be represented by the indicator function of a set, i.e. $y=f^*(F)$ for some $\{0,1\}$-valued function $F\in\mathcal F.$
By the Krein-Milman theorem, the set $(f^*)^{-1}(y)\cap\mathcal F$ has an extreme point $F.$ I claim that $F$ is the indicator function of a set.
Suppose for contradiction that $\{x\mid 0<F(x)<1\}$ has positive measure. Let $X_1,\dots,X_{n+1}$ be disjoint positive measure subsets of $\{x\mid 0<F(x)<1\},$ using the fact that $[0,1]$ is non-atomic.
Define $D_\lambda(x)=\sum_{i=1}^{n+1}\lambda_i \min(F(x),1-F(x)) 1_{X_{i}}(x)$ for $\lambda\in\mathbb R^{n+1}.$ Note $f^*(D_\lambda)$ is linear in $\lambda.$ Since $f^*$ has rank at most $n$ there is some $\lambda\in[-1,1]^{n+1}\setminus\{0\}$ such that $f^*(D_\lambda)=0,$ giving $F=\tfrac12((F+D_\lambda)+(F-D_{\lambda})).$ Since $F+D_\lambda$ and $F-D_\lambda$ are both in $(f^*)^{-1}(y)\cap\mathcal F$ and distinct from $F,$ this proves that $F$ is not an extreme point, contradicting the choice of $F.$
A: This is an answer for the first version of the question, where we had $f: {\mathbb R}^n\rightarrow {\mathbb R}$
For your second statement, assuming $f\in L^1(E)$, consider for $t\in {\mathbb R}$
$$\phi(t) = \int_E f(x)\chi_{x_1\le t} d\mu$$
then $\phi$ is continuous in ${\mathbb R}$ by the dominated convergence theorem and $\lim_{t\to-\infty}\phi(t) = 0$ and $\lim_{t\to+\infty}\phi(t) = \int_E f(x) d\mu$, so that $\phi$ reaches any value between $0$ and $\int_E f d\mu$.
Edit Still supposing that $f \in L^1(E)$, I realize that the above argument applies to any measurable subset $A\subset E$, so that the whole interval $\left[0, \int_A f d\mu\right]$ belongs to the set $C =\{\int_X f d\mu,X\subset E\}$. It follows that $C$ is a union of intervals that all contain $0$ and therefore it is an interval containing $0$.
With very little more work, one sees that
$$C = \left[-\int_E f^- d\mu, \int_E f^+d\mu\right]$$
where $f^- = \max(0, -f)$ and $f^+ = \max(0,f)$.
QED
A: This is an attempt to answer the second version of the question with $f: [0, 1] \rightarrow {\mathbb R}^n$.
If $f$ is piecewise constant, i.e.~if $f$ takes only $N$
values ${V}_{1} , \cdots  , {V}_{N}$, let ${m}_{k}$ be the measure
of ${f}^{{-1}} ({V}_{k})$, then obviously for any subset $A$, the
function $f \left(t\right) {{\chi}}_{A} \left(t\right)$ will take the same values, and one has
$$\int_{A}^{}f d t = \sum _{i = 1}^{N} {{\alpha}}_{k} {V}_{k}$$
where ${{\alpha}}_{k} \in  \left[0 , {m}_{k}\right]$. Every such sum is reachable,
so that the reachable set is the convex set
$$C = \left\{\sum _{i = 1}^{N} {{\alpha}}_{k} {V}_{k} , \quad  \forall  k , {{\alpha}}_{k} \in  \left[0 , {m}_{k}\right]\right\}\qquad (1)$$
For non piecewise constant function, one could build an approximation of the
reachable set by the following construction: partition ${\mathbb{R}}^{n}$ in a sequence
of disjoint cubes of side ${\epsilon}$
$${\mathbb{R}}^{n} = \bigcup_{j \in  \mathbb{N}} {K}_{j}^{{\epsilon}}$$
and let ${V}_{j}^{{\epsilon}}$ be the center of the cube ${K}_{j}^{{\epsilon}}$.
We approximate $f$ by a piecewise constant function ${f}_{{\epsilon}}$ by
defining
$${f}_{{\epsilon}} \left(t\right) = {V}_{j}^{\epsilon} \quad  \text{if} \  f \left(t\right) \in  {K}_{j}^{{\epsilon}}$$
Then formula (1) defines a convex set ${C}^{{\epsilon}}$ approximating the set $C$
of reachable values.
This proof is not complete of course, it remains to prove some kind of convergence
result for the obtained family ${C}^{{\epsilon}}$.
