Hi guys I need help proving the equivalence of the following. I don't know how to do that. Given are a probability space $(\Omega, \mathcal{A}, P)$, a filtration $(\mathcal{F}_n)_{n\in\Bbb{N}\cup\{\infty\}} $with $\mathcal{F}_\infty := \sigma(\bigcup_{n\in\Bbb{N}} \mathcal{F}_n) = \mathcal{A}$ and another probability measure $Q$ over $(\Omega , \mathcal{A})$.
For $n \in\Bbb{N}$ we designate $P_n = P|_{\mathcal{F}_n}$ resp. $Q_n = Q|_{\mathcal{F}_n}$ the restriction of $P$ or $Q$ on $\mathcal{F}_n$. For each $n \in\Bbb{N}:\;Q_n$ has a density $M_n : \Omega → \Bbb{R}_+^0$ with respect to $P_n$. Furthermore $M_\infty :=\lim \inf_{n\to\infty} M_n : \Omega → \Bbb{R}_+^0$ Now I need to prove the equivalence of these statements.
(a) $M_\infty$ is a density of $Q$ with respect to $P$.
(b) $Q$ has a density with respect to $P$.
(c) The stochastic process $(M_n)_{n\in\Bbb{N}}$ on $(\Omega, \mathcal{A}, P)$ is uniformly integrable.
(d) The stochastic process $(M_n)_{n\in\Bbb{N} \cup \{\infty\}} $ is a Martingale on the filtered probability space $(\Omega, \mathcal{A}, P,(\mathcal{F}_n)_{n\in\Bbb{N} \cup \{\infty\}})$.
(e) $E_P [M_\infty] =1$.
I'm going to assume that you're familiar with Doob's convergence theorem about martingales.
Let $(\Omega,\mathcal A,\mathbb P)$ be a probabilistic space, and let $(\mathcal F_n)_{n \in \mathbb N}$ be a filtration. Let $F_{\infty}:= \sigma(\bigcup_{n=1}^\infty \mathcal F_n)=\mathcal A$ and let $Q$ be another probability measure on space $(\Omega,\mathcal A)$. By $\mathbb P_n,Q_n$ denote restrictions of respectivelly $\mathbb P,Q$ to sigma-fields $\mathcal F_n$. Let's assume that $Q_n$ has density $M_n$ with respect to $\mathbb P_n$, that is for $A \in \mathcal F_n$ we get $Q_n(A) = \int_{A}M_n(\omega)d\mathbb P_n(\omega)$.
Before we start, let's note that $(M_n,\mathcal F_n)_{n \in \mathbb N}$ is a martingale on $(\Omega,\mathcal A,\mathbb P)$. Indeed, via definition of density, $M_n$ is $\mathcal F_n$ measurable. Moreover $\mathbb E_{\mathbb P}[M_n] = \int_{\Omega}M_n(\omega)d\mathbb P(\omega) = \int_{\Omega}M_n(\omega)d\mathbb P_n(\omega) = 1$ since $\Omega \in \mathcal F_n$ and $M_n$ is a density with respect to $\mathbb P_n$. For the last part, we need to check $\mathbb E_{\mathbb P}[M_{n+1} | \mathcal F_n] = M_n$ a.s. So take any $A \in \mathcal F_n$. This means that $A \in \mathcal F_{n+1}$, too. Then we have $$ \int_A M_{n+1} d\mathbb P = \int_A M_{n+1}d\mathbb P_{n+1} = Q_{n+1}(A) = Q_n(A) = \int_AM_nd\mathbb P_n = \int_A M_n d\mathbb P$$ So that $(M_n,\mathcal F_n)_{n \in \mathbb N}$ is indeed a martingale on $(\Omega,\mathcal A,\mathbb P)$. By martingale convergence theorem (we have non-negative martingales), there exists almost surely finite random variable $M_{\infty} = \lim_n M_n$.
Now we can proceed with question.
Firstly, let's show that $(a)$ is equivalent with $(b)$.
Implication $(a) \to (b)$ is trivial, so let's show $(b) \to (a)$. Assume that $Y$ is a density of $Q$ with respect to $\mathbb P$. This in particular means that for any $n \in \mathbb N$, $A \in \mathcal F_n$ we get $$ \int_A Yd\mathbb P = Q(A) = Q_n(A) = \int_A M_n d\mathbb P_n = \int_A M_n d\mathbb P$$ This means that we get $\mathbb E_{\mathbb P}[Y|\mathcal F_n] = M_n$, but by Doob's theorem, martingale with such a form converges almost surely too $\mathbb E_{\mathbb P}[Y | \mathcal F_{\infty}] = Y$. So we have simultaneously that $M_n \to M_{\infty}$ almost surely and $M_n \to Y$ almost surely, hence $M_{\infty}$ is a density function of $Q$ what had to be proved.
Secondly note that $(c)$ and $(d)$ are equaivalent.
Indeed note that by Doob's convergence in $L_1$ theorem, uniform integrability is equivalent with $M_n \to M_{\infty}$ in $L_1$ which is again equivalent with statement $\mathbb E[M_{\infty}|\mathcal F_n] = M_n$ and the last one (together with $(M_{n},\mathcal F_n)_{n \in \mathbb N}$ is a martingale on $(\Omega,\mathcal A,\mathbb P)$) gives us that $(M_n,\mathcal F_n)_{n \in \mathbb N \cup \{\infty\}}$ is a martingale on $(\Omega,\mathcal A,\mathbb P)$ so that indeed $(c)$ and $(d)$ are equivalent.
Thirdly $(a),(b),(c),(d)$ are equivalent.
Indeed, from $(a) \to (d)$, by looking at our proof in "Firstly", we showed that $M_n = \mathbb E_{\mathbb P}[M_\infty | \mathcal F_n]$ and it's enought (together with $(M_n,\mathcal F_n)_{n \in \mathbb N}$ being a martingale and $M_{\infty}$ being density so measurable and integrable) to say that $(M_n,\mathcal F_n)_{n \in \mathbb N \cup \{\infty\}}$ is a martingale on $(\Omega,\mathcal A,\mathbb P)$.
Other way around, from $(d) \to (a)$. $(d)$ means that for any $n \in \mathbb N, A \in \mathcal F_n$ we have $$ Q(A) = \int_{A}M_n d\mathbb P = \int_A M_{\infty} d\mathbb P$$ Note that set $\mathcal L := \{ A \in \mathcal A : Q(A)= \int_A M_{\infty} d\mathbb P \}$ is a $\lambda - $ system ( you can check by simple properties of measures and integrals like for $A \subset B$ : $Q(B \setminus A) = Q(B) - Q(A) , \int_{B \setminus A} = \int_B - \int_A$ and for $A_1 \subset A_2 \subset ...$ you have $Q(\bigcup A_k) = \lim_k Q(A_k)$ and (if integrating integrable random variable) $\int_{\bigcup A_k} = \lim_k \int_{A_k}$ (by dominated convergence). Note that $\mathcal C := \bigcup \mathcal F_n$ is a $\pi-$system such that $\mathcal C \subset \mathcal L$ so by monotone class theorem/Dynkin's lemma we get $\sigma(\mathcal C) \subset \mathcal L$ which means $\mathcal L = \mathcal A$ and we're done.
Lastly we need to join $(e)$ together with the rest. Implication $(a) \to (e)$ is trivial. I've finally managed to show $(e) \to (c)$ with some help.
We need to show that for any $\varepsilon > 0$ there exists $K$ such that $\sup_n \mathbb E[M_n 1_{M_n > K}] < \varepsilon.$
So fix $\varepsilon > 0$. For fixed $K > 0$ define function $f_K : \mathbb R_+ \to \mathbb R_+$ by formula $$f_K(x) = \begin{cases} x & x \in [0,K] \\ 0 & x > K+1 \\ continuous \in [0,K] & x \in (K,K+1] \end{cases}$$ (by continuos $\in [0,K]$ I meant that we can have whatever values such that $f_K$ is continuous with values in $[0,K]$). There exists such $K$ that $ \mathbb E[M_{\infty}] - \mathbb E[f_K(M_{\infty})] < \varepsilon $ (by dominated convergence, since $M_{\infty}$ is integrable and $f_K(x) \le x$). Now we have $K$ fixed (but note that if we take $K^* > K$ then $\mathbb E[M_{\infty}] - \mathbb E[f_{K^*}(M_{\infty})] < \varepsilon$ also holds). Now, by $M_n \to M_{\infty}$ almost surely and boundedness/continuity of $f_K$ (and again dominated convergence) we have $N$ such that for $n \ge N$ it holds : $ |\mathbb E[f_K(M_n)] - \mathbb E[f_K(M_{\infty})]| < \varepsilon$
Hence for $n \ge N$ we get using assumption ($\mathbb E[M_n] = \mathbb E[M_{\infty}]$): $$ \mathbb E[M_n 1_{M_n > K}] \le \mathbb E[M_n] - \mathbb E[f_K(M_n)] \le \mathbb E[M_{\infty}] + \varepsilon - \mathbb E[f_K(M_{\infty})] < 2\varepsilon$$
What's left, is to enlarge $K$ to $K^*$ such that the above holds for $n \in \{1,...,N-1\}$, too (we can do it, since we're left with only finitelly many random variables, and for every $n \in \{1,...,N-1\}$ we can find such $K_n$ that $\mathbb E[M_n 1_{M_n > K_n}] < 2\varepsilon$. Then take $K^* = \max\{K,K_1,...,K_{N-1}\}$ and we're done.