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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.

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  • $\begingroup$ Why is the first statement equivalent to the second one? $\endgroup$ Aug 31, 2017 at 12:19
  • $\begingroup$ Is $f$ assumed positive? in $L^1$? $\endgroup$
    – H. H. Rugh
    Aug 31, 2017 at 13:04
  • $\begingroup$ $f$ is defined on $[0,1]$ and takes values in $R^n$. Sorry about the confusion. $\endgroup$
    – Kitegi
    Aug 31, 2017 at 15:16
  • $\begingroup$ The question makes no sense to me. What statement is supposed to be equivalent to that other thing? $\endgroup$
    – zhw.
    Aug 31, 2017 at 16:15
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    $\begingroup$ @Gribouillis For a subset of length $\alpha$, the second term is $\alpha$, and the first term can be any number between $\alpha^2/2$ and $\alpha(1-\alpha/2)$. So the set of "reachable" values is $\{(x,y) \text{ s.t. } y^2/2\leq x\leq y(1-y/2)\}$, which is convex since the lower bound is convex and the upper bound is concave. $\endgroup$
    – Kitegi
    Aug 31, 2017 at 19:04

3 Answers 3

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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 $F.$

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  • $\begingroup$ Sorry for taking so long to respond. I've been a bit busy lately and haven't found a good chance to carefully read your proof. I've put a new bounty since the previous one ended up running out. I should be able to award it after 24 hours. $\endgroup$
    – Kitegi
    Apr 19, 2018 at 16:41
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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

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  • $\begingroup$ I made a mistake when stating the question. What I need to prove is something different. $\endgroup$
    – Kitegi
    Aug 31, 2017 at 15:20
  • $\begingroup$ @Fannight Well, start another thread then! $\endgroup$ Aug 31, 2017 at 15:49
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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 ${C}^{{\epsilon}}$.

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  • $\begingroup$ I had a similar approach, but the part I had trouble with is the convergence of the family of sets. I'm not sure if there's a convenient way to say that a family of sets converges. I think it would be easier to do if we had more properties about the set of reachable values. I think it's closed in $R^n$ for all functions but I can't seem to prove it. $\endgroup$
    – Kitegi
    Aug 31, 2017 at 20:45
  • $\begingroup$ You could start with a regular $f$. For such a function, the approximation by the piecewise constant function is uniform. You could try to prove that if $x \in C$, then $d(x,C_\epsilon)$ tends to 0 for example. You could say that $x$ is in the limit set if every neighborhood of $x$ intersects all the $C^\epsilon$ when $\epsilon$ is small enough, etc. There are many options. $\endgroup$ Aug 31, 2017 at 20:53

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