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\[\int\chi_{E}\,dx\geq\sup_{n}\int\chi_{E\cap[-n,n]}\,dx=\sup_{n}m(E\cap[-n,n])=m (E)=\infty.\]
**Proposition 38**.: _The integral of non-negative measurable functions satisfies the following properties:_
* _Linearity. If_ \(f,g\geq 0\)_, and_ \(a,b\) _are positive real numbers, then_ \[\int(af+bg)\,dx=a\int f\,dx+b\int g\,dx.\]
* _Additivity. If_ \(E\) _and_ \(F\) _are disjoint subsets of_ \(\mathbb{R}\)_, and_ \(f\geq 0\)_, then_ \[\int_{E\cup F}f\,dx=\int_{E}f\,dx+\int_{F}f\,dx.\]
* _Monotonicity. If_ \(0\leq f\leq g\) _then_ \[\int f\,dx\leq\int g\,dx.\]
* _If_ \(g\) _is integrable and_ \(0\leq f\leq g\)_, then_ \(f\) _is integrable as well._
* _If_ \(f\) _is integrable, then_ \(f(x)<\infty\) _for almost every_ \(x\)_._
* _If_ \(\int f\,dx=0\)_, then_ \(f(x)=0\) _for almost every_ \(x\)_._
Proof
: We only give proofs of parts e) and f). Parts b), c), and d) follow from the definitions; part a) is not immediate but we leave to the reader.
To prove part e), set up the sets \(E_{k}=\{x:\,f(x)\geq k\}\) and \(E_{\infty}=\{x:\,f(x)=\infty\}\). From the set \(E_{k}\) we can write
\[k\,m(E_{k})\leq\int\chi_{E_{k}}f\,dx\leq\int f\,dx\]
and hence \(m(E_{k})\to 0\) as \(k\to\infty\). Since \(E_{k}\) decreases to \(E_{\infty}\), using the continuity property of measures we obtain \(m(E_{\infty})=\lim_{k\to\infty}m(E_{k})=0\).
To prove f), if \(f\geq 0\), supported on a set of finite measure \(E\) and \(\int f\,dx=0\) then we want to show \(f=0\) almost everywhere. For each integer \(k\geq 1\) define \(E_{k}=\{x\in E:\ f(x)\geq 1/k\}\); then from the inequality \(k^{-1}\chi_{E_{k}}\leq f\) and the monotonicity of the integral we see that
\[m(E_{k})\leq\frac{1}{k}\int f\,dx.\]
Thus \(m(E_{k})=0\) for all \(k\), and since
\[\{x:\ f(x)>0\}=\bigcup_{k=1}^{\infty}E_{k},\]
we obtain the desired result that \(f=0\) almost everywhere.
### General Lebesgue Integral
Suppose \(f\) is a measurable function. Then we divide \(f\) into its positive and negative parts as follows:
\[f^{+}(x)=\sup(f(x),0)\qquad f^{-}(x)=\sup(-f(x),0)=-\inf(f(x),0).\]
**Definition 99**.: A measurable function \(f\) is _integrable_ if the non-negative measurable function \(|f|\) is integrable in the sense of previous section.
That is, if \(\int\mid f\mid\,dx<\infty\).
Note that the functions \(f^{\pm}\) satisfy \(0\leq f^{\pm}\leq\mid f\mid\), so \(f\) is integrable if and only if \(\int f^{+}\,dx<\infty\) and \(\int f^{-}\,dx<\infty\).
**Definition 100**.: If \(f\) is integrable then its integral is
\[\int f\,dx=\int f^{+}\,dx-\int f^{-}\,dx.\]
This is consistent with the earlier definition, since \(f\geq 0\) implies \(f^{+}=f\) and \(f^{-}=0\).
**Theorem 117** (Basic Properties of the Integral).: _The integral of Lebesgue integrable functions is linear, additive, monotonic and satisfies the triangle inequality._
Proof
: These properties require proof, but we only prove additivity, since \((f+g)^{\pm}\neq f^{\pm}+g^{\pm}.\) However if \(f\) and \(g\) are integrable, so is \(f+g\) since we have the triangle inequality \(\mid f+g\mid\leq\mid f\mid+\mid g\mid\). Next observe that
\[(f+g)^{+}-(f+g)^{-}=f+g=(f^{+}-f^{-})+(g^{+}-g^{-}).\]
Therefore
\[(f+g)^{+}+f^{-}+g^{-}=(f+g)^{-}+f^{+}+g^{+}.\]All of the functions in the above identity are non-negative so we can apply additivity to obtain
\[\int(f+g)^{+}\,dx+\int f^{-}\,dx+\int g^{-}\,dx=\int(f+g)^{-}\,dx+\int f^{+}\,dx +\int g^{+}\,dx,\]
and so
\[\int(f+g)^{+}\,dx-\int(f+g)^{-}\,dx=\int(f+g)\,dx=\int(f^{+}-f^{-})\,dx+\int(g^ {+}-g^{-})\,dx\]
yielding
\[\int(f+g)\,dx=\int f\,dx+\int g\,dx.\qed\]
**Remark 64**.: If \(f\) is integrable on \(\mathbb{R}\), \(E\) is a measurable set and \(f\) is defined on \(E\), the integral of the restriction over \(E\) is defined as \(\int f\chi_{E}\,dx\), and \(f\chi_{E}=0\) on \(E^{c}\), the complement of \(E\). When defining the product \(f\chi_{E}\) we use \(0(\pm\infty)=0\).
**Remark 65** (Tchebychev Inequality).: Suppose \(f\geq 0\) and integrable. If \(\lambda>0\) and \(E_{\lambda}=\{x:\ f(x)>\lambda\}\), then one can show
\[m(E_{\lambda})\leq\frac{1}{\lambda}\int f\,dx.\]
The importance of this inequality is that it estimates the “size” of \(f\) in terms of the integral of \(f\).
### The Space \(L^{1}\) of Integrable Functions
The set of integrable functions form a vector space, so it is appropriate to ask how one can place a norm on it so it becomes a complete normed space.
**Definition 101**.: For any integrable function \(f\) on \(\mathbb{R}\), we define the \(L^{1}\)- _norm_ of \(f\) as:
\[||\,f\,||=||\,f\,||_{L^{1}}=\int_{\mathbb{R}}|f(x)|\,dx.\]
Note that \(||\,f\,||=0\) if and only if \(f=0\) a.e.. Furthermore, we do not distinguish between two functions that agree almost everywhere; with these facts in mind we define \(L^{1}(\mathbb{R})\) as the set of all equivalence classes of integrable functions (two functions \(f\) and \(g\) are equivalent if \(f=g\) a.e.). We list the elementary facts about the norm \(||\,f\,||_{L^{1}}\) in the following proposition.
**Proposition 39**.: _Let \(f,g\in L^{1}(\mathbb{R})\), then_
* \(||\,f\,||_{L^{1}}=0\) _if and only if_ \(f=0\) _a.e.._
* \(||\,kf\,||_{L^{1}}=|k|||\,f\,||_{L^{1}}\) _for all scalars_ \(k\)_._
* \(||\,f+g\,||_{L^{1}}\leq||\,f\,||_{L^{1}}+||\,g\,||_{L^{1}}\)This norm induces a metric \(d(x,y)=||\,f-g\,||_{L^{1}(\mathbb{R})}\) on \(L^{1}(\mathbb{R})\). There are two properties of \(L^{1}(\mathbb{R})\) we now mention without proof (see [50], p. 70-71):
* The vector space \(L^{1}(\mathbb{R})\) is complete in its metric (called the _Riesz-Fisher Theorem_).
* The families of simple functions, step functions, and continuous functions with compact support are dense in \(L^{1}(\mathbb{R})\).
### Convergence Theorems
In this section we see three fundamental convergence theorems in Lebesgue integration theory. Namely the dominated convergence theorem (DCT), monotone convergence theorem (MCT), and Fatou’s lemma. In all of these theorems we have a sequence of functions \((f_{n})\) such that \(f_{n}\to f\) a.e. Then Fatou’s lemma, MCT, and DCT all state under suitable conditions that we can assert something about \(\int f\,dx\) in terms \(\int f_{n}\,dx\). Fatou’s lemma has the weakest hypothesis; we need to only have \(f_{n}\) bounded below by \(0\). Consequently its conclusion is weaker; we can only conclude
\[\int f\,dx\leq\liminf\int f_{n}\,dx.\]
In DCT, we assume \(f_{n}\) is bounded above and below by an integrable function \(g\), i.e., \(|\;f_{n}(x)\;| \[\int f\,dx=\lim\int f_{n}\,dx.\] In MCT we require \(f_{n}\) to be bounded below by zero and bounded above by the limit function \(f\). In other words we assume \(0\leq f_{1}\leq f_{2}\leq\cdots\leq f\) and conclude \[\int f\,dx=\lim\int f_{n}\,dx.\]
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