So I am trying to attain the solution to Mathematical Statistics: Exercises and Solutions by Jun Shao Chapter $1$, Exercise $15$:
Let $P$ and $Q$ be two probability field measures on a $\sigma$-field $\mathcal{F}$. Assume that $f=\dfrac{dP}{dv}$ and $g=\dfrac{dQ}{dv}$ exists on a measure $v$ on $\mathcal{F}$. Show that$$\int\vert f-g\vert dv=2\operatorname{sup}\{\vert P(C)-Q(C)\vert:C\in\mathcal{F}\}$$
However, since my recent question was not exactly well received, I decided to try and do this without being allowed the following:
- My previous question (which could possibly help me in some way)
- Any online resources
- I cannot use any calculators
- I will also have to include the restriction that I am forced to upload a picture of all of my work that I did on paper at the end of my proof, to (hopefully) avoid any confusion of me possibly cheating. I am not denying any possibility that I did, in fact, cheat when creating the proof to the other exercise, I am only saying that it is possible that I did cheat, and that is why I am putting the restrictions for creating the proof to this exercise.
$$\mathbf{\text{My proof}}$$
Letting $A=\{f\geq g\}$ and $B=\{f\lt g\}$, which implies that $C\cap A,C\cap B$ for any $C\in\mathcal{F}$
$$\implies A,B\in\mathcal{F}$$
$$\implies\int\vert f-g\vert dv=\int_A(f-g)dv+\int_B(g-f)dv$$
$$\therefore=P(A)-Q(A)+P(B)-Q(B)\text{, which is a reference to the original statement}$$
$$\text{since }\int\vert f-g\vert dv=2\operatorname{sup}\{\vert\color{red}{P(C)-Q(C)}\vert:C\in\mathcal{F}\}$$$$\text{which this part can be broken up as }P(A)-Q(A)+P(B)-Q(B)$$
$$\implies\leq\vert P(A)-Q(A)\vert+\vert P(B)-Q(B)\vert$$
$$\text{which is because the absolute value of a number is always}$$$$\text{is always greater than or equal to the original value}$$$$\text{i.e. }\operatorname{abs}(3)=3\text{, }\operatorname{abs}(-3)=3$$$$\therefore\text{ }-3,3\leq3$$$$\therefore[-3,3]\text{ is in the set of }\mathbb{R}\leq3$$
$$\implies\leq2\operatorname{sup}\{\vert P(C)-Q(C)\vert:C\in\mathcal{F}\}$$
$$\text{Since it is implied that }\vert P(A)-Q(A)\vert+\vert P(B)-Q(B)\vert\sim\vert P(C)-Q(C)\vert$$
$$\therefore\text{ For any }C\in\mathcal{F}$$$$\int_C(f-g)dv=\int_{C\cap A}(f-g)dv+\int_{C\cap B}(f-g)dv\leq\int_A(f-g)dv$$
$$\therefore P(C)-Q(C)=\int_{C^c}(g-f)dv\leq\int_B(g-f)dv$$
$$\therefore\implies2[\vert P(C)-Q(C)\vert]\leq\int\vert f-g\vert dv$$
$$\therefore\int\vert f-g\vert dv\geq2\operatorname{sup}\{\vert P(C)-Q(C)\vert:C\in\mathcal{F}\}$$$$\text{Which in turn, proves our original statement in the sense that}$$
$$\int\vert f-g\vert dv\geq2\operatorname{sup}\{\vert P(C)-Q(C)\vert:C\in\mathcal{F}\}$$$$\sim\int\vert f-g\vert dv=2\operatorname{sup}\{\vert P(C)-Q(C)\vert:C\in\mathcal{F}\}$$$$\color{white}{doafwrwerq}\overset{\large\uparrow}{\text{Our original statement that we are trying to prove}}$$
A photo of my proof:
(sorry if it looks horrible/unreadable, IMHO (in my honest opinion) that might be why photos/screenshots probably aren't allowed on the MSE other than it won't turn up/turn up horribly on other devices/browsers)
$$\mathbf{\text{My question}}$$
Is the proof that I have formulated correct, or what could I do to formulate it correctly/formulate it more easily?