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# 数学代写|勒贝格积分代写Lebesgue Integration代考|TMA4225 General Measurable Functions

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## 数学代写|勒贝格积分代写Lebesgue Integration代考|General Measurable Functions

In this section we return to Lebesgue measure $\mu$ on the interval $[0,1]$ but, consider extended measurable functions which may be unbounded both above and below. We define
$$f^{+}(x)=\max {f(x), 0}$$
and
$$f^{-}(x)=-\min {f(x), 0} .$$
These are both non-negative measurable functions and
$$f(x)=f^{+}(x)-f^{-}(x) .$$
Definition 4.4.1. (Lebesgue integrable). If $f:[0,1] \rightarrow[-\infty, \infty]$ is a measurable function, then we say $f$ is Lebesgue integrable provided both $f^{+}$and $f^{-}$are integrable (as non-negative functions). If $f$ is integrable, we define
$$\int f d \mu=\int f^{+} d \mu-\int f^{-} d \mu .$$
Proposition 4.4.2. Suppose $f$ and $g$ are measurable functions on $[0,1]$ and $f=g$ almost everywhere. Then if $f$ is integrable, so is $g$ and $\int f d \mu=\int g d \mu$. In particular, if $f=0$ almost everywhere, $\int f d \mu=0$.

Proof. If $f$ and $g$ are measurable functions on $[0,1]$ and $f=g$ almost everywhere, then $f^{+}=g^{+}$almost everywhere, $f^{-}=g^{-}$almost everywhere, and $f^{+}$and $f^{-}$are integrable. It then follows from Proposition 4.1.4 that $g^{+}$and $g^{-}$are integrable and that $\int f^{+} d \mu \geq$ $\int g^{+} d \mu$ and $\int f^{-} d \mu \geq \int g^{-} d \mu$. Switching the roles of $f$ and $g$ this same proposition gives the reverse inequalities, so we have $\int f^{+} d \mu=\int g^{+} d \mu$ and $\int f^{-} d \mu=\int g^{-} d \mu$.

## 数学代写|勒贝格积分代写Lebesgue Integration代考|Convergence in L2

We have discussed uniform convergence and pointwise convergence and now we wish to discuss convergence in the $L^2[a, b]$ norm || . Note that again we are adopting the customary convention by which || is called a norm even though $|f|=0$ implies only that $f(x)=0$ almost everywhere. So again if we were to be pedantic this is a norm on the vector space of equivalence classes of functions which are equal almost everywhere. With this caveat the vector space $L^2[a, b]$ is a metric space with distance function given by $\operatorname{dist}(f, g)=|f-g|$.
Definition 5.2.1. (Convergence in $L^2$ ). If $\left{f_n\right}_{n=1}^{\infty}$ is a sequence in $L^2[a, b]$, then it is said to converge to $f$ in $L^2[a, b]$ if
$$\lim _{n \rightarrow \infty}\left|f-f_n\right|=0 .$$
Functions which are bounded form a dense subset of $L^2[a, b]$. In fact, the following result shows that a good bounded function approximating $f \in L^2[a, b]$ can be obtained by taking $n$ large and defining $f_n(x)$ to be $n$ if $f(x)>n$, or $-n$ if $f(x)<-n$, and $f(x)$ otherwise.

## 数学代写|勒贝格积分代写lebesgue积分代考|一般可测函数

$$f^{+}(x)=/max f(x), 0$$

$$f（x）=f^{+}（x）-f^{-}（x） 。$$

$$\int f d \mu=int f^{+} d \mu-int f^{-} d \mu$$

## 数学代写|勒贝格积分代写lebesgue积分代写|l2中的收敛性

$lim _{n \rightarrow \infty}\left|f-f_n\right|=0$

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