# 数学代写|泛函分析作业代写Functional Analysis代考|MAST90020

If $f: \Omega \rightarrow \mathbb{K}$ is integrable with respect to the measure $\mu$, then the $\mathbb{K}$-valued measure
$$v(F):=\int_F f \mathrm{~d} \mu, \quad F \in \mathscr{F},$$
is absolutely continuous with respect to $\mu$, that is, $\mu(F)=0$ implies $v(F)=0$. The following theorem provides a converse under a $\sigma$-finiteness assumption.

Theorem 2.46 (Radon-Nikodým). Let $(\Omega, \mathscr{F}, \mu)$ be a $\sigma$-finite measure space. If the measure $v: \mathscr{F} \rightarrow \mathbb{K}$ is absolutely continuous with respect to $\mu$, then there exists a unique $g \in L^1(\Omega, \mu)$ such that
$$v(F)=\int_F g \mathrm{~d} \mu, \quad F \in \mathscr{F} .$$
Proof Uniqueness being clear, the proof is devoted to proving existence. By considering real and imaginary parts separately it suffices to consider the case of real scalars. Then, decomposing $v$ into positive and negative parts via the Jordan decomposition, it suffices to consider the case where $v$ is a finite nonnegative measure.
Consider the set
$$S:=\left{f \in L^1(\Omega, \mu): f \geqslant 0, \int_F f \mathrm{~d} \mu \leqslant v(F) \text { for all } F \in \mathscr{F}\right} .$$
Then $0 \in S$, so $S$ is nonempty. Let
$$M:=\sup {f \in S} \int{\Omega} f \mathrm{~d} \mu .$$
For all $f \in S$ we have $\int_{\Omega} f \mathrm{~d} \mu \leqslant v(\Omega)$ and therefore $M \leqslant v(\Omega)<\infty$.
Step 1 – In this step we prove that there exists a function $g \in S$ for which the supremum in the definition of $M$ is attained. Let $\left(f_n\right)_{n \geqslant 1}$ be a sequence in $S$ with the property that $\lim {n \rightarrow \infty} \int{\Omega} f_n \mathrm{~d} \mu=M$. Set $g_n:=f_1 \vee \cdots \vee f_n$. Any set $F \in \mathscr{F}$ can be written as a disjoint union of sets $F_1^{(n)}, \ldots, F_n^{(n)} \in \mathscr{F}$ such that $g_j=f_j$ on $F_j^{(n)}$ and therefore
$$\int_F g_n \mathrm{~d} \mu=\sum_{j=1}^n \int_{F_j^{(n)}} f_j \mathrm{~d} \mu \leqslant \sum_{j=1}^n v\left(F_j^{(n)}\right)=v(F)$$

## 数学代写|泛函分析作业代写Functional Analysis代考|Integration with Respect to K-Valued Measures

A measurable function $f$ is said to be integrable with respect to a $\mathbb{K}$-valued measure $\mu$ if it is integrable with respect to $|\mu|$. The function $f$ is integrable with respect to a real measure $\mu$ if and only if it is integrable with respect to the measures $\mu^{+}$and $\mu^{-}$, where $\mu=\mu^{+}-\mu^{-}$is the Jordan decomposition, and $f$ is integrable with respect to a complex measure $\mu$ if and only if $f$ is integrable with respect to the real and imaginary parts of $\mu$.

The integral of an integrable function $f$ with respect to a real measure $\mu$ is defined by
$$\int_{\Omega} f \mathrm{~d} \mu:=\int_{\Omega} f \mathrm{~d} \mu^{+}-\int_{\Omega} f \mathrm{~d} \mu^{-},$$
and the integral of an integrable function $f$ with respect to a complex measure $\mu$ by
$$\int_{\Omega} f \mathrm{~d} \mu:=\int_{\Omega} f \mathrm{~d} \operatorname{Re} \mu+i \int_{\Omega} f \mathrm{~d} \operatorname{Im} \mu .$$
Proposition 2.49. If $f$ is integrable with respect to a $\mathbb{K}$ valued measure $\mu$, then
$$\left|\int_{\Omega} f \mathrm{~d} \mu\right| \leqslant \int_{\Omega}|f| \mathrm{d}|\mu| .$$
Proof First let $f=\sum_{n=1}^N c_n \mathbf{1}{F_n}$ be a simple function, with the sets $F_n \in \mathscr{F}$ disjoint. Then $$\left|\int{\Omega} f \mathrm{~d} \mu\right|=\left|\sum_{n=1}^N c_n \mu\left(F_n\right)\right| \leqslant \sum_{n=1}^N\left|c_n\right|\left|\mu\left(F_n\right)\right| \leqslant \sum_{n=1}^N\left|c_n\right||\mu|\left(F_n\right)=\int_{\Omega}|f| \mathrm{d}|\mu| .$$
The general case follows from this by observing that the simple functions are dense in $L^1(\Omega,|\mu|)$ and that $f_n \rightarrow f$ in $L^1(\Omega,|\mu|)$ implies $\int_{\Omega}\left|f_n-f\right| \mathrm{d} v \rightarrow 0$ for each of the measures $v \in\left{\operatorname{Re} \mu, \operatorname{Im} \mu, \mu^{+}, \mu^{-}\right}$.

A more elegant, but less elementary, alternative definition of the integral $\int_{\Omega} f \mathrm{~d} \mu$ can be given with the help of the Radon-Nikodým theorem. Indeed, defining $\int_{\Omega} f \mathrm{~d} \mu$ as above, by the result of Example $2.48$ for functions $f \in L^1(\Omega,|\mu|)$ we have the identity
$$\int_{\Omega} f \mathrm{~d} \mu=\int_{\Omega} f h \mathrm{~d}|\mu|,$$
where $\mathrm{d} \mu=h \mathrm{~d}|\mu|$ as in the example (note that $f h \in L^1(\Omega,|\mu|$ ) since $|h|=1 \mu$-almost everywhere). This identity could be taken as an alternative definition for the integral $\int_{\Omega} f \mathrm{~d} \mu$

# 泛函分析代考

$$v(F):=\int_F f \mathrm{~d} \mu, \quad F \in \mathscr{F},$$

$$v(F)=\int_F g \mathrm{~d} \mu, \quad F \in \mathscr{F} .$$

$$M:=\sup f \in S \int \Omega f \mathrm{~d} \mu .$$

## 数学代写|泛函分析作业代写Functional Analysis代考|Integration with Respect to K-Valued Measures

$$\int_{\Omega} f \mathrm{~d} \mu:=\int_{\Omega} f \mathrm{~d} \mu^{+}-\int_{\Omega} f \mathrm{~d} \mu^{-},$$

$$\int_{\Omega} f \mathrm{~d} \mu:=\int_{\Omega} f \mathrm{~d} \operatorname{Re} \mu+i \int_{\Omega} f \mathrm{~d} \operatorname{Im} \mu .$$

$$\left|\int_{\Omega} f \mathrm{~d} \mu\right| \leqslant \int_{\Omega}|f| \mathrm{d}|\mu| .$$

$$\left|\int \Omega f \mathrm{~d} \mu\right|=\left|\sum_{n=1}^N c_n \mu\left(F_n\right)\right| \leqslant \sum_{n=1}^N\left|c_n\right|\left|\mu\left(F_n\right)\right| \leqslant \sum_{n=1}^N\left|c_n\right||\mu|\left(F_n\right)=\int_{\Omega}|f| \mathrm{d}|\mu| .$$

$\int_{\Omega}\left|f_n-f\right| \mathrm{d} v \rightarrow 0$ 对于每一项措施

$$\int_{\Omega} f \mathrm{~d} \mu=\int_{\Omega} f h \mathrm{~d}|\mu|,$$

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