📂Measure Theory

Lebesgue Integration

Buildup

Before considering the generalization of Riemann integration, it is necessary to define what a simple function is.

Let’s say the range $\phi : \mathbb{R} \to \mathbb{R}$ of the function values, which are non-negative, is a finite set $\left\{ a_{1} , a_{2}, \cdots , a_{n} \right\}$. If it satisfies $A_{i} = \phi^{-1} \left( \left\{ a_{i} \right\} \right) \in \mathcal{M}$, then $\phi$ is called a simple function. Simple functions have the following properties:

• (i): If $i \ne j$, then $A_{i } \cap A_{j} = \emptyset$
• (ii): $\displaystyle \bigsqcup_{k=1}^{n} A_{k} = \mathbb{R}$
• (iii): $\displaystyle \phi (x) = \sum_{k=1}^{n} a_{k} \mathbb{1}_{A_{k}}(x)$ is a measurable function.

Simple functions are, by their very definition, composed of three elements that are too easy to handle. Firstly, since the function values are non-negative, there is no need to consider signs; secondly, being finite makes addition and subtraction flexible; and thirdly, they are measurable. The word simple is used in various ways in different fields of mathematics, but at least in real analysis, it can be considered the opposite of ‘complex’. Having defined simple functions that are easy and convenient to handle, it is immediately possible to think of a new integration that covers Riemann integration.

Lebesgue Integration of Simple Functions

When $\phi$ is a simple function and $E \in \mathcal{M}$, $\displaystyle \int_{E} \phi dm := \sum_{k=1}^{n} a_{k} m (A_{k} \cap E)$ is called the Lebesgue Integral of the simple function $\phi$. The Lebesgue integral has the following properties:

• [1]: For all $a>0$, $\displaystyle \int_{E} a \phi dm = a \int_{E} \phi dm$
• [2]: For two simple functions $\phi , \psi$, if $\phi \le \psi$, then $\displaystyle \int_{E} \phi dm \le \int_{E} \psi dm$
• [3]: If $A, B \in \mathcal{M}$ for $A \cap B = \emptyset$, then $\displaystyle \int_{A \cup B} \phi dm = \int_{A} \phi dm + \int_{B} \phi dm$

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• $m$ is a Lebesgue measure.

However, the condition of being a simple function is too strong and specific to be widely applicable. Adding an idea like the method of exhaustion roughly completes a satisfactory ‘Lebesgue Integral’.

Definition ¹

When $\phi$ is a simple function, for a non-negative measurable function $f$ and $E \in \mathcal{M}$, $$\displaystyle \int_{E} f dm := \sup \left\{ \left. \int_{E} \phi dm \ \right| \ 0 \le \phi \le f \right\}$$ is called the Lebesgue Integral of the measurable function $f$.

Fundamental Properties

The Lebesgue integral has the following properties:

• [1]’: For all $r \ge 0$, $\displaystyle \int_{E} r f dm = r \int_{E} f dm$
• [2]’: For two simple functions $f, g$, if $f \le g$, then $\displaystyle \int_{E} f dm \le \int_{E} g dm$
• [3]’: If $A, B \in \mathcal{M}$ for $A \cap B = \emptyset$, then $\displaystyle \int_{A \cup B} f dm = \int_{A} f dm + \int_{B} f dm$
• [4]’: If $A, B \in \mathcal{M}$ for $A \subset B$, then $\displaystyle \int_{A} f dm \le \int_{B} f dm$
• [5]’: If $N \in \mathcal{N}$, then $\displaystyle \int_{N} f dm = 0$
• [6]’: $\displaystyle m(E) \inf_{E} f \le \int_{E} f dm \le m(E) \sup_{E} f$

Explanation

In addition to these basic properties, the following theorem can be considered. Using this theorem, calculations as novel as $\displaystyle \int_{\mathbb{R}} \mathbb{1}_{\mathbb{Q}} dm = 0$ can be completed in a single cut. Though not as simple to prove as it appears, it’s certainly worth a look at least once.

Theorem

For measurable functions $f \ge 0$ in a measurable space $( X , \mathcal{E} )$ and all measurable sets $A \in \mathcal{E}$, $$\int_{A} f dm = 0 \iff f = 0 \text{ a.e.}$$

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• $\text{a.e.}$ means almost everywhere.

Proof

$( \implies )$

If $E := f^{-1} ( 0 , \infty)$ for $m(E) = 0$, then $f$ is almost everywhere $f=0$. Assuming $\displaystyle E_{n} := f^{-1} \left[ {{1} \over {n}} , \infty \right)$, if $\displaystyle E = \bigcup_{n=1}^{\infty} E_{n}$ while $\displaystyle \lim_{n \to \infty} E_{n} = E$ holds true. Considering the simple function $\displaystyle \phi_{n} := {{1}\over {n}} \mathbb{1}_{E_{n}} \le f$, we have $${{1}\over {n}} m( E_{n} ) = \int_{A} \phi_{n} dm \le \int_{A} f dm = 0$$ Therefore, $${{1} \over {n}} m(E_{n}) \le 0$$ meaning, for all $n \in \mathbb{N}$, $m(E_{n}) = 0$.

[7]: $E_{n} \in \mathcal{M}$, $\displaystyle E_{n} \subset E_{n+1} \implies m \left( \bigcup_{n=1}^{\infty} E_{n} \right) = \lim_{n \to \infty} m (E_{n})$

Meanwhile, since $E_{n} \subset E_{n+1}$, the following holds true: $$m \left( \bigcup_{n=1}^{\infty} E_{n} \right) = \lim_{n \to \infty} m (E_{n}) = m(E) = 0$$

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$( \impliedby )$

Since $f$ is almost everywhere $f=0$ and the simple function $\phi$ satisfies $0 \le \phi \le f$, $\phi$ is also almost everywhere $\phi = 0$. Thus, $\displaystyle \int_{A} f dm = 0$ is true.

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