Prove that the set of reachable values is convex.

Question

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.

Answer 1

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

Answer 2

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 ${C}^{{\epsilon}}$.