Let $E \subseteq [0,1]$, $\mu$ the Lebesgue measure.

I would like to show that $\int_{E} x \, d\mu \geq \frac{1}{2} \mu(E)^2$.

Lemma: $$ \int_{0}^{\mu(E)} x \, d\mu \leq \int_{E} x \, d\mu $$

This lemma seems pretty reasonable, in fact, I would expect it to hold for any monotone function $f : [0, 1] \to \mathbb{R}$:

$$ \int_{0}^{\mu(E)} f \, d\mu \leq \int_{E} f \, d\mu $$

But I'm not sure how to prove it rigorously. If the lemma holds, the result follow as $\int_{0}^{\mu(E)} x \, d\mu = \frac{1}{2} \mu(E)^2$.

Idea of a proof:

\begin{align*} \int_{0}^{1} 1_{E} f \, d\mu &= \left[ f(x) \int_{0}^{x} 1_{E} \, d\mu \right]_{0}^{1} - \int_{0}^{1} \left( \int_{0}^{x} 1_{E} \, d\mu \right) f'(x) \, d\mu(x) \\ &= f(1) \mu(E) - \int_{0}^{1} \left( \int_{0}^{x} 1_{E} \, d\mu \right) f'(x) \, d\mu(x) \\ &\geq f(1) \mu(E) - \int_{0}^{1} \left( \int_{0}^{x} 1_{[0, \mu(E)]} \, d\mu \right) f'(x) \, d\mu(x) \\ &= \int_{0}^{1} 1_{[0, \mu(E)]} f \, d\mu \end{align*}

Again, I'm not sure if integration by parts is valid in this context, and I'm pretty sure that $f$ needn't be differentiable for the lemma to hold.

  • $\begingroup$ I'm pretty sure that $\int_0^a x\,dx = a^2/2$. Where did the $1/2$ go? $\endgroup$ Aug 5, 2016 at 19:59
  • $\begingroup$ Hate to break it to you, but is this true? Let $E=[0,1].$ Then $\int_E x d\mu = 1/2$ but $\mu(E)^2=1$ $\endgroup$ Aug 5, 2016 at 20:00
  • $\begingroup$ Sorry, I lost a factor of two, updated now $\endgroup$
    – user357269
    Aug 5, 2016 at 20:03

4 Answers 4


Here's an elementary solution. Note that
$$ \begin{aligned} \frac 12 \mu(E)^2 &= \int x1_{[0,\mu(E)]}(x) d\mu(x)\\ &= \int x1_{[0,\mu(E)]\cap E}(x) d\mu(x) + \int x1_{[0,\mu(E)]\cap E^c}(x) d\mu(x)\\ &= \int x1_{E}(x) d\mu(x)- \int x1_{[0,\mu(E)]^c\cap E}(x) d\mu(x) + \int x1_{[0,\mu(E)]\cap E^c}(x) d\mu(x)\\ &= \int x1_{E}(x) d\mu(x)- \int x1_{[\mu(E),1]\cap E}(x) d\mu(x) + \int x1_{[0,\mu(E)]\cap E^c}(x) d\mu(x)\\ &\leq \int x1_{E}(x) d\mu(x) - \mu(E) \int 1_{[\mu(E),1]\cap E}(x) d\mu(x) + \mu(E)\int 1_{[0,\mu(E)]\cap E^c}(x) d\mu(x)\\ &= \int x1_{E}(x) d\mu(x) - \mu(E) \mu([\mu(E),1]\cap E) + \mu(E)\mu([0,\mu(E)]\cap E^c)\\ &= \int x1_{E}(x) d\mu(x) +\mu(E)[\mu([0,\mu(E)])-\mu([0,\mu(E)]\cap E)-\mu([\mu(E),1]\cap E)]\\ &= \int x1_{E}(x) d\mu(x) +\mu(E)[\mu(E) - \mu(E)]\\ &= \int x1_{E}(x) d\mu(x) \end{aligned} $$


Here's an idea to prove the lemma: We can approximate $E$ by an open set of the form $O=\bigcup_{i=1}^n(a_i,b_i)$, with $a_1<b_1<a_2<\cdots<b_n$. Then on one hand, we have $\mu(E)\sim\sum_{i=1}^n(b_i-a_i)$, and then $$\int_0^{\mu(E)}xd\mu(x)\sim\int_0^{\sum_{i=1}^n{b_i-a_i}}xd\mu(x)=\sum_{i=1}^n\int_{c_i}^{c_{i+1}} xd\mu(x)$$ where $c_1=0$, and $c_{i+1}=(b_1-a_1)+\cdots+(b_i-a_i)$. On the other hand, we have $$\int_E xd\mu(x)\sim\sum_{i=1}^n\int_{a_i}^{b_i}xd\mu(x)$$ Now we can show, say by induction, that for each $i$, the interval $(c_i,c_{i+1})$ stands to the left of the interval $(a_i,b_i)$, and since the identity function is increasing, we have $\int_{c_i}^{c_{i+1}}xd\mu(x)\leq\int_{a_i}^{b_i}xd\mu(x)$. Summing this over all $i$, we obtain $$\int_0^{\mu(E)}xd\mu(x)\leq\int_E xd\mu(x),$$ and the same argument works in the case of a non-decreasing function.

You can make these approximation arguments formal with an $\epsilon$-argument.

  • $\begingroup$ Thanks, but this is a proof for the case when $E$ is a finite disjoint union of intervals, which is not really the generality that I'm interested in $\endgroup$
    – user357269
    Aug 5, 2016 at 23:45
  • $\begingroup$ @user357269 In the second and last phrases: "We can approximate $E$ by an open set of the form..." and "You can make these arguments formal with an $\epsilon$-argument", I mean you should fill all the details where we use the "$\sim$" sign, which means "approximately equal", in this case, by starting with some $\epsilon>0$ and verifying approximate identities. All the functinsb involved (namely, the identity) are bounded, so these approximations can be done. I'm sorry if this was not clear, but it is common practice when giving simply the idea of the proof (as stated in the first phrase). $\endgroup$ Aug 6, 2016 at 4:19

This isn't a complete answer, but it was pointed out to me that this is a special case of the Hardy-Littlewood inequality.


Write $G(x) = \mu(E\cap[0,x]) $ and let $f : [0, 1] \to \mathbb{R}$ be non-decreasing. By noting that $x \geq G(x)$, we get

$$ \int_{0}^{1} f(x) \mathbf{1}_{E}(x) \, \mathrm{d}x = \int_{0}^{1} f(x) \, \mathrm{d}G(x) \geq \int_{0}^{1} f(G(x)) \, \mathrm{d}G(x) = \int_{0}^{G(1)} f(u) \, \mathrm{d}u. $$

If OP is not familiar to the theory of Lebesgue-Stieltjes integral, we can adopt an approximation argument to circumvent this technicality in the following way.

Let $g : [0, 1] \to [0, 1]$ be continuous and define $G(x) = \int_{0}^{x} g(t) \, \mathrm{d}t$. Then for $f$ as before,

$$ \int_{0}^{1} f(x) g(x) \, \mathrm{d}x = \int_{0}^{1} f(x) G'(x) \, \mathrm{d}x \geq \int_{0}^{1} f(G(x))G'(x) \, \mathrm{d}x = \int_{0}^{G(1)} f(u) \, \mathrm{d}u, $$

Now, since we know that the space $C([0,1])$ of continuous functions is dense in $L^1([0,1])$, we can find a sequence $(g_n)_{n\geq 1}\subset C([0,1])$ such that $g_n \to \mathbf{1}_E$ in $L^1$. Since $f$ is bounded, this implies

\begin{align*} \int_{0}^{1} f(x)\mathbf{1}_{E}(x) \, \mathrm{d}x &= \lim_{n\to\infty} \int_{0}^{1} f(x)g_n(x) \, \mathrm{d}x \\ &\geq \lim_{n\to\infty} \int_{0}^{\|g\|_{L^1}} f(x) \, \mathrm{d}x = \lim_{n\to\infty} \int_{0}^{\|\mathbf{1}_E\|_{L^1}} f(x) \, \mathrm{d}x. \end{align*}

Since $\|\mathbf{1}_E\|_{L^1} = \mu(E)$, the desired conclusion follows.


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