Lebesgue measurable function and set, $\forall t\in (0,1), $there exists $ E_t \subset E$ such that $\int_{E_t} f(x) dx = t \int_{E} f(x) dx$.

Question

The question is: Let f be Lebesgue measurable on $R$ and $E\subset R$ be measurable so that $0<A=\int _E f(x)dx < \infty$. Show that for every $t\in (0,1)$ there exists a measurable set $E_t \subset E$ so that $\int_{E_t} f(x) dx = t A$.

My thoughts to solve this question and my question is in step 3:

1. Since $f$ is Lebesgue measurable on $R$, so $f$ can be approximated by simple functions. So if I can show the result is true for $f(x) = \chi _{B}(x)$ where $\chi _{B}(x)$ is a simple function. Then the result is true for Lebesgue measurable $f$.

2. Now $\int _{E} \chi _{B}(x) dx = m(E\cap B)= A$ ($m$ is Lebesgue measure). Now I have to show that there exists a measurable set $E_t \subset E$ such that $m(E_t\cap B)=t \cdot m(E\cap B)= tA$.

3. To show that $m(E_t\cap B)=t \cdot m(E\cap B)= tA$, here is the trouble for me if the above steps are correct. After searching posts here, I think I can define a $g(x) = m(E\cap B\cap (-\infty, x))$. I need to show $g(x)$ is continuous and then that could lead to the conclusion that there will be $t_0$ such that $0<g(t_0) = t \cdot m(E\cap B)<m(E\cap B)$.

I would like to hear about your comments about the above steps and see if there is any error. Thanks!

Answer 1

For your statement to make sense, $f$ must be assumed to be not only measurable but integrable (but $f$ is not necessarily of constant sign, and its integral $A\in\Bbb R$ does not need to be $>0$). Assume moreover wlog that $f$ is $0$ outside $E.$

Your last sentence in step 1 is unproved, and useless. Instead, consider the function $$g:s\mapsto\int_{-\infty}^sf(x)\,dx.$$ The continuity of $g$ is a standard result, quite easy to prove.

Since $\lim_{-\infty}g=0$ and $\lim_{+\infty}g=A,$ by the intermediate value theorem $$\forall t\in(0,1)\quad\exists s\in\Bbb R\quad g(s)=tA,$$ and $E_t:=(-\infty,s)\cap E$ is then a solution.

Answer 2

Regarding your step 1, I'm afraid I'm not clear on how the general version of this result would follow from proving this result for simple functions. It's true that for any measurable $f$, there exists a sequence of simple functions $s_n$ that converge to $f$ pointwise. However, an $E_t$ that "works" for one $s_n$ might not "work" for another $s_n$...

Your idea for step 3, however, is very nice. We can take that idea and apply it to the general case, not just to the special case where the function is simple.

Concretely, the idea is to consider the function $g : \mathbb R \to \mathbb R$ defined by $$ g(x) := \int_{E \cap (-\infty, x)} f(u) du. $$ You want to prove that:

• $g$ is continuous at all $x \in \mathbb R$.
• $\lim_{x \to -\infty} g(x) = 0$
• $\lim_{x \to +\infty} g(x) = A$.

To prove this, I suggest you use the sequential definition of continuity and limits. That is, I suggest you prove that $$\lim_{n \to \infty} x_n = l \ \ \implies \ \ \lim_{n \to \infty} \int_{E \cap (-\infty, x_n)} f(u) du = \int_{E \cap (-\infty, l)} f(u) du.$$ (Here, $l$ could be in $\mathbb R$, or $l$ could be $-\infty$ or $+\infty$.) The Dominated Convergence Theorem should get you most of the way there - think about the pointwise limit of the sequence of functions $f \mathbf 1_{E \cap (-\infty, x_n)}$, and bear in mind that this sequence is dominated by the positive integrable function $|f| \mathbf 1_E$.

Finally, having done all this work, you should be able to obtain your result by applying the intermediate value theorem to $g$.

Answer 3

Since your $f$ can take both positive and negative values on $E$, I'll break up the solution to two steps.

Step 1: Reduce to the case where $f$ on its domain of integration is a nonnegative measurable function.

On $E$, let $f^+ = \max(0,f)$ and $f^- = \max(0,-f)$, so $f^+$ and $f^-$ are positive on disjoint measurable subsets $E^{+}$ and $E^{-}$ of $E$, with $f = f^+ - f^{-}$ on $E$. Then $$ \int_E f(x)\,dx = \int_{E} f^+(x)\,dx - \int_{E} f^{-}(x)\,dx = \int_{E^{+}} f^+(x)\,dx - \int_{E^{-}} f^{-}(x)\,dx. $$ Suppose your result can be shown when the Lebesgue measurable functions being integrated are nonnegative. Then apply this to $f^+$ on $E^+$ and $f^-$ on $E^-$: for $t \in (0,1)$ there are measurable subsets $E^+_t \subset E^+$ and $E^{-}_t \subset E^{-}$ such that $$ \int_{E^+_t}f^+(x)\,dx = t\int_{E^+} f^+(x)\,dx = t\int_{E^+} f(x)\,dx, \ \ \ \int_{E^{-}_t}f^-(x)\,dx = t\int_{E^-} f^-(x)\,dx = t\int_{E^-} (-f(x))\,dx. $$ Since $E^+_t$ and $E^{-}_t$ are inside the disjoint subsets $E^+$ and $E^{-}$, $$ \int_{E^+_t \cup E^{-}_t} f(x)\,dx = \int_{E^+_t} f(x)\,dx + \int_{E^{-}_t} f(x)\,dx = \int_{E^+_t} f^+(x)\,dx - \int_{E^{-}_t} f^{-}(x)\,dx $$ and $$ \int_{E^+_t} f^+(x)\,dx - \int_{E^{-}_t} f^{-}(x)\,dx = t\int_{E^+} f(x)\,dx + t\int_{E^-} f(x)\,dx = t\int_E f(x)\,dx, $$ so you can use for $E_t$ the subset $E_t^+ \cup E_t^{-}$.

Step 2: Prove the case where $f \geq 0$ on a domain of integration $E$.

The condition $\int_E f(x)\,dx < \infty$ makes integration of $f$ on measurable subsets of $E$ a finite measure, say $\nu$: define $\nu(A) := \int_A f(x)\,dx$ for measurable subsets $A$ in $E$. Moreover, $\nu$ is nonatomic.

Now we can appeal to the following general theorem about values of finite nonatomic measures.

Theorem. Let $(X,\nu)$ be a finite nonatomic measure space. For each $t$ in $[0,\nu(X)]$, there is a measurable subset $X_t \subset X$ such that $\nu(X_t) = t$.

Proof. See Theorem 4.2 here.