# How to prove an inequality of Lebesgue integral?

Definition of measurable set: A set $E$ is measurable if $$m^*(T) = m^*(T \cap E) + m^*(T \cap E^c)$$ for every subset of $T$ of $\mathbb R$.

Definition of Lebesgue measurable function: Given a function $f: D \to \mathbb R ∪ \{+\infty, -\infty\}$, defined on some domain $D \subset \mathbb{R}^n$, we say that $f$ is Lebesgue measurable if $D$ is measurable and if, for each $a\in[-\infty, +\infty]$, the set $\{x\in D \mid f(x) > a\}$ is measurable.

Definition of Lebesgue integral of simple function: We say that a simple function $\psi$ is Lebesgue integrable if the set $\{\psi \ne 0\}$ has finite measure. In this case, we may write the standard representation for $\psi$ as $\psi = \sum_{i=0}^n a_i \chi_{A_i}$, where $a_0 = 0, a_1, \ldots , a_n$ are distinct real numbers, where $A_0 = \{\psi = 0\}, A_1, \ldots , A_n$ are pairwise disjoint and measurable, and where only $A_0$ has infinite measure, Once $\psi$ is so written, there is an obvious definition for $\int \psi$, namely, $$\int \psi = \int_{\mathbb R} \psi = \int_{-\infty}^{+\infty} \psi(x) \ \mathsf dx = \sum_{i=1}^n a_i m(A_i).$$ In other words, by adopting the convention that $0 \cdot \infty = 0$, we define the Lebesgue integral of $\psi$ by $$\int \sum_{i=0}^n a_i \chi_{A_i} = \sum_{i=0}^n a_i m(A_i).$$ Please note that $a_im(A_i)$ is a product of real numbers for $i \ne 0$, and it is $0 \cdot \infty = 0$ for $i = 0$; that is, $\int \psi$ is a finite real number.

Definition of Lebesgue integrable of non-negative function: If $f: \mathbb R \to [0, +\infty]$ is measurable, we define the Lebesgue integral of $f$ over $\mathbb R$ by $$\int f = \sup \left\{\int \psi: 0 \le \psi \le f, \psi\text{ simple function and integrable }\right\}.$$

Suppose $f(x)$ and $g(x)$ are non-negative Lebesgue measurable functions defined on $E \subset \mathbb R$. How to prove $$\left(\int_{E} f(x)\ \mathsf dx\right)^{\frac{1}{2}} \left(\int_{E} g(x)\ \mathsf dx\right)^{\frac{1}{2}} \ge \int_{E} {\sqrt{f(x)g(x)}}\ \mathsf dx$$?

Besides, can anyone introduce some mathematical books about inequalities to me such as basic inequalities, integral inequalities, norm inequalities, convex inequalities and so on? I'm not sure whether there is a book covering such a board area of mathematical inequalities. So it will be pretty helpful even if your introduced book merely covering one specific area of them. Thanks in advance.

Update:

I notice that if I treat $a = \sqrt {f(x)}$ and $b = \sqrt {g(x)}$ then it will be much looked like a Holder inequality in Riemann integral.

• Look carefully - there's an extra $1/2$ in your formula. As long as we're talking about non-math, note what happens if you say \left(\int_E\right) instead of (\int_E). Anyway, this is exactly Holder's inequality. You've evidently seen a proof for the Riemann integral - the same proof works here. (At least I can't imagine a proof for the Riemann integral that doesn't work for the Lebesgue integral; check back if you find a proof for the Riemann integral but have problems adapting it...) – David C. Ullrich Aug 5 '15 at 22:36
• Note every \left( has to have a matching \right) or it won't work... – David C. Ullrich Aug 5 '15 at 22:36
• @DavidC.Ullrich: Sorry, my typo. – Bear and bunny Aug 5 '15 at 22:40
• @DavidC.Ullrich: Sorry, I can't figure out what's the difference between \left(\int_E\right) and (\int_E). I typed in and latex returned correspondingly two almost same symbols(actually, exactly same in my eyes). Can you explain more? – Bear and bunny Aug 5 '15 at 22:50
• @Bearandbunny I edited it in. \left and \right make parentheses/brackets/braces/etc. resize to match the height of their contents. – Math1000 Aug 5 '15 at 23:30

When $||f||_p > 0, ||g||_{p'}>0$ and $p, p' < +\infty$, then$$a^{\frac{1}{p}}b^{\frac{1}{p'}} \le \frac{a}{p} + \frac{b}{p'}, a>0, b>0$$ and $a = \frac{|f(x)|^p}{||f||_{p}^{p}}, b = \frac{|g(x)|^{p'}}{||g||_{p'}^{p'}}$, then get $$\frac{|f(x)g(x)|}{||f||_p ||g||_{p'}} \le \frac{1}{p}\frac{|f(x)|^p}{||f||_p^p} + \frac{1}{p'} \frac{|g(x)|^{p'}}{||g||_{p'}^{p'}}$$. Integrate both side, then $$\int_E \frac{|f(x)g(x)|}{||f||_p ||g||_{p'}}dx \le \frac{1}{p} \int_E \frac{|f(x)|^p}{||f||_p^p}dx + \frac{1}{p'} \int_E \frac{|g(x)|^{p'}}{||g||_{p'}^{p'}}dx = \frac{1}{p} + \frac{1}{p'} = 1$$ where $$\int_E {|f(x)|^p} dx =||f||_p^p, \int_E {|g(x)|^{p'}} dx =||g||_{p'}^{p'}$$.