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Suppose we want to find
\[\sum_{k=1}^{n}(y_{k+1}-y_{k})\ (\text{length of the interval}\,\{x:f(x)\geq y_{k}\}).\]
However, the set \(\{x:f(x)\geq y_{k}\}\) might be a complicated set and one asks how can one find its length? To answer this question one has to develop the idea of measure (or length) of a set. That is why we introduced the concept of measure first. Among many notions of integral the most prominent one is the Lebesgue integral. To develop these ideas precisely requires a text in itself (see [8, 22, 50]). However, here we can answer some basic questions. Lebesgue integral defined by a step by step process, and in each step one is able to integrate larger families of functions. We define Lebesgue integral of simple functions first, then we move on to bounded functions supported on a set of finite measure, then the Lebesgue integral of non-negative functions is introduced before we go to general case which is the class of integrable functions. We start with the building blocks for the theory of integration, namely the characteristic function \(\chi_{E}\) of \(E\) (sometimes called the indicator function of \(E\)) which is
\[\chi_{E}(x)=\left\{\begin{array}{ll}1&\text{if}\,\ x\in E\\ 0&\text{if}\,\ x\notin E\end{array}\right..\]
It is easy to see that the height of \(\chi_{E}\) is one; therefore, the area under the function \(\chi_{E}\) must be the length of \(E\), or more precisely \(m(E)\). We denote this by \(\int_{E}\chi_{E}\,dx.\) This implies that the area under \(a\chi_{E}\) is then equal to
\[a\,m(E)=\int_{E}a\,\chi_{E}\,dx.\]
Now consider two disjoint sets \(E\) and \(F\), with the same arguments the area under the function \(a\,\chi_{E}+b\,\chi_{F}\) must be equal to \(a\,m(E)+b\,m(F)\), and we write this as (Figure 2.27):
\[\int_{E\cup F}(a\,\chi_{E}+b\,\chi_{F})\,dx=a\,m(E)+b\,m(F).\]
Figure 2.26: Dividing by horizontal rectangles
### The Lebesgue Integral of Simple Functions
Suppose \(\varphi\) is a simple function in the canonical form \(\sum_{k=1}^{N}a_{k}\chi_{E_{k}}\) where the \(E_{k}\) are measurable sets of finite measure and the \(a_{k}\)’s are distinct and nonzero, (canonical form assures that decomposition \(\varphi\) into sum is unique). In this case we define the _Lebesgue integral_ of \(\varphi\) by
\[\int_{\mathbb{R}}\varphi\,dx=\sum_{k=1}^{N}a_{k}m(E_{k}).\]
If \(F\) is a measurable subset of \(\mathbb{R}\) with finite measure, we can define the Lebesgue integral of \(\varphi\) over \(F\) by
\[\int_{F}\varphi\,dx=\int\varphi(x)\chi_{F}(x)\,dx.\]
We simply use \(\int\varphi\,dx\) for the integral of \(\varphi\) over \(\mathbb{R}\). If clarity requires the variable to be emphasized, then the notation \(\int\varphi(x)\,dx\) will be used. Note that if we need to use representations of \(\varphi\) which are not canonical then we need to prove \(\int_{\mathbb{R}}\varphi\,dx=\sum_{k=1}^{N}a_{k}m(E_{k})\) is independent of any representation of \(\varphi\). For the proof of this claim see [50], p. 51.
The integral of simple functions defined above satisfies the following properties:
* If \(\varphi\) and \(\psi\) are simple and \(a,b\in\mathbb{R}\), then \[\int(a\varphi+b\psi)\,dx=a\int\varphi\,dx+b\int\psi\,dx\quad\text{(Linearity)}.\]
* If \(E\) and \(F\) are disjoint subsets of \(\mathbb{R}\) with finite measure, then \[\int_{E\cup F}\varphi\,dx=\int_{E}\varphi\,dx+\int_{F}\varphi\,dx\quad\text{ (Additivity)}.\]* If \(\varphi\leq\psi\) are simple, then \[\int\varphi\,dx\leq\int\psi\,dx\quad\text{(Monotonicity)}.\]
* If \(\varphi\) is a simple function, then so is \(|\varphi|\), and \[\left|\int\varphi\right|\,dx\leq\int|\varphi|\,dx\quad\text{ (Triangle Inequality)}.\]
##### The Lebesgue Integral of a Bounded Function over a Set of Finite Measure
**Definition 97**.: Let \(f\) be a measurable function, then the _support_ of \(f\) is the set of points \(x\) for which \(f(x)\neq 0\). In other words, \(\operatorname{supp}(f)=\{x:\ f(x)\neq 0\}\). Note that if \(f\) is measurable so is the set \(\operatorname{supp}(f)\).
Recalling the theorem of approximating measurable functions, Theorem 112, if a function \(f\) is supported on a set \(E\) and bounded say by a constant \(M\), then
* There exists a sequence of simple functions \((\varphi_{n})\) such that \(\varphi_{n}(x)\to f(x)\) for all \(x\).
* Each \(\varphi_{n}\) is bounded on \(E\) by \(M\).
* Each \(\varphi_{n}\) is supported on \(E\).
The importance of these cases stems from Egorov’s result (see Theorem 115) which states that the convergence of a sequence of measurable functions is “nearly” uniform. Now we give the definition of integration for bounded functions that are supported on a set of finite measure. For such a function we define its Lebesque integral by:
\[\int f(x)\,dx=\lim_{n\to\infty}\int\varphi_{n}(x)\,dx,\]
where \(\{\varphi_{n}\}\) is any sequence of simple functions satisfying the conditions \(a),b)\), and \(c)\) stated above. However we do not know if this limit exists. Egorov’s theorem can be applied here. Since the measure of \(E\) is finite, given \(\epsilon>0\) there exists a closed measurable subset \(E_{\epsilon}\) of \(E\) such that \(m(E\setminus E_{\epsilon})\leq\epsilon\) and \(\varphi_{n}\to f\) uniformly on \(E_{\epsilon}\). Therefore by setting \(I_{n}=\int\varphi_{n}\,dx\), we can estimate \(|I_{n}-I_{m}|\) as
\[|I_{n}-I_{m}|\leq\int_{E_{\epsilon}}|\varphi_{n}-\varphi_{m}|\,dx+\int_{E \setminus E_{\epsilon}}|\varphi_{n}-\varphi_{m}|\,dx\leq\int_{E_{\epsilon}}| \varphi_{n}-\varphi_{m}|\,dx+2M\epsilon.\]
By the uniform convergence, we have, for all \(x\in E_{\epsilon}\) and for all sufficiently large \(n\) and \(m\), the estimate \(|\varphi_{n}-\varphi_{m}|<\epsilon\), so we conclude that
\[|I_{n}-I_{m}|\leq m(E)\epsilon+2M\epsilon\]for all large \(n\) and \(m\). Since \(m(E)<\infty\) and \(\epsilon\) is arbitrary, this shows that \(\{I_{n}\}\) is a Cauchy sequence and hence converges. Therefore \(\lim\limits_{n\to\infty}\int\varphi_{n}(x)\,dx\) exists. Furthermore, if \(f=0\) a.e., then repeating the argument above one can show \(|I_{n}|\leq(m(E)+2M)\epsilon\) which yields \(\lim\limits_{n\to\infty}\int\varphi_{n}(x)\,dx=0\). It is not difficult to show that \(\int f\,dx\) is independent of the limiting sequence \(\varphi_{n}\) used in order for the integral to be well defined. (Why?) Furthermore if \(f\) is a simple function itself, then \(\int f\,dx\) as defined above coincides with the integral of simple functions given before. Thus one can extend the basic properties of integration defined for simple functions, to bounded functions supported on a set of finite measure.
**Proposition 37**.: _Suppose \(f\) and \(g\) are bounded functions supported on a set of finite measure. Then the following properties hold._
* _If_ \(a,b\in\mathbb{R}\)_, then_ \[\int(af+bg)\,dx=a\int f\,dx+b\int g\,dx\quad\text{(Linearity)}.\]
* _If_ \(E\) _and_ \(F\) _are disjoint subsets of_ \(\mathbb{R}\) _with finite measure, then_ \[\int_{E\cup F}f\,dx=\int_{E}f\,dx+\int_{F}g\,dx\quad\text{(Additivity)}.\]
* _If_ \(f\leq g\)_, then_ \[\int f\,dx\leq\int g\,dx\quad\text{(Monotonicity)}.\]
* _If_ \(f\) _is bounded then_ \(|f|\) _is also bounded, supported on a set of finite measure, and_ \[\left|\int f\,dx\right|\leq\int|f|\,dx\quad\text{(Triangle Inequality)}.\]
All of the properties follow directly from using approximation of measurable functions by simple functions, and the properties of the integral of simple functions were given previously.
For bounded functions supported on a set of finite measure we have a crucial convergence theorem. This theorem is about the interchange of an integral and a limit. Because its conclusion is simply:
\[\lim\limits_{n\to\infty}\int f_{n}\,dx=\int\lim\limits_{n\to\infty}f_{n}\,dx.\]
**Theorem 116** (Bounded Convergence Theorem (Bct)).: _Suppose \((f_{n})\) is a sequence of measurable functions such that \(|\ f_{n}(x)\ |\leq M\) for all \(n\) and \((f_{n})\) supported on a set \(E\) of finite measure. Suppose also \(f_{n}(x)\to f(x)\) a.e. as \(n\to\infty\). Then the limit function \(f\) is measurable, bounded, supported on \(E\) for a.e. \(x\), and_
Proof
: By Egorov’s Theorem 115, we know there is measurable subset \(E_{\epsilon}\) of \(E\) such that \(m(E\setminus E_{\epsilon})\leq\epsilon\) and \(f_{n}\to f\) uniformly on \(E_{\epsilon}\). Thus for all sufficiently large \(n\) we must have
\[\mid f_{n}(x)-f(x)\mid\leq\epsilon\quad\text{for all}\quad x\in E_{\epsilon}.\]
Using the properties of the integral,
\[\int\mid f_{n}-f\mid\;dx\leq\int_{E_{\epsilon}}\mid f_{n}-f\mid\;dx+\int_{E \setminus E_{\epsilon}}\mid f_{n}-f\mid\;dx\leq\epsilon\,m(E)+2Mm(E\setminus E _{\epsilon})\]
for all \(n\) large enough. Since \(\epsilon\) is arbitrary, the proof is completed.
Observe from \(\int\mid f_{n}-f\mid\;dx\to 0\) as \(n\to\infty\) we conclude \(\int f_{n}\,dx\to\int f\,dx\) as \(n\to\infty\). This theorem will be developed further in the section on “convergence theorems.” Another useful observation at this point is the fact that if \(f\geq 0\), bounded and a supported on a set of finite measure \(E\) and \(\int f\,dx=0\) then \(f=0\) almost everywhere. For the proof see part \(f)\) of Proposition 38 below.
**Remark 63**.: If we assume \(f\) is Riemann integrable on a closed interval \([a,b]\), then \(f\) is measurable, and the standard Riemann integral, is equal to its Lebesgue integral. In other words
\[(R)\int_{[a,b]}f\,dx=(L)\int_{[a,b]}f\,dx,\]
where the integral on the left-hand side is the Riemann Integral, and the one on the right-hand side is the Lebesgue integral. (See the exercises at the end of this section). This result combined with bounded convergence theorem shows why one can integrate much larger class of functions in the sense of Lebesgue.
### The Lebesgue Integral of Non-Negative Functions
We next define the integral of non-negative measurable but not necessarily bounded functions. We will allow these functions to be extended-valued, i.e., these functions may take the value \(+\infty\) on a measurable set. It is useful to remember the convention that one defines the supremum of a set of positive numbers to be \(+\infty\) if the set is unbounded.
**Definition 98**.: Let \(f:\mathbb{R}\to[0,\infty]\) be measurable, we define the _Lebesgue integral_ of \(f\) over \(\mathbb{R}\) by
\[\int f(x)\,dx=\sup_{g}\int g(x)\,dx,\]
where the supremum is taken over all measurable functions \(g\) such that \(g\) is bounded and supported on a set of finite measure and \(0\leq g\leq f\).
In the above definition of the integral, there are two possible cases: the supremum is either finite or infinite, if \(\int f\,dx<\infty\), then we say \(f\) is _Lebesgue integrable_ on \(\mathbb{R}\). It follows that if \(E\) is a measurable subset of \(\mathbb{R}\), and \(f\geq 0\) then \(f\chi_{E}\) is also positive and we define
\[\int_{E}f(x)\,dx=\int f(x)\chi_{E}(x)\,dx.\]
Note that we are not excluding the possibility that \(\int f(x)\,dx=\infty\). In particular, if \(E\) is measurable and \(m(E)<\infty\) then as usual \(m(E)=\int\chi_{E}\,dx\). However when \(m(E)=\infty\), we have
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