Uniform integrable proof

Question

Lets be $E[\sup_{t\in[0,T]}e^{-rt}\Psi(S_{t})]<\infty$. I want to show that $J_{t}$ defined by \begin{align} J_{t}:=\mathrm{ess sup}_{\tau \in \mathcal F_{t,T}}E[e^{-r\tau}\Psi(S_{\tau})|\mathcal F_{t}] \end{align} for some stopping time $\tau$ is of Class D. So i have to show $\{J_{\nu}|\nu \in \mathcal F_{0,T}\}$is uniformly integrable. I found a proof like this: \begin{align} J_{\nu}\leq E[\sup_{t\in[0,T]}e^{-rt}\Psi(S_{t})|\mathcal F_{\nu}] \end{align} For $\epsilon>0: \exists\delta >0, A\in \mathcal F_{T}: P(A)<\delta$ such that \begin{align} \int_{A}\sup_{t\in[0,T]}e^{-rt}\Psi(S_{t})dP<\epsilon \end{align} Choose $\alpha>E[\sup_{t\in[0,T]}e^{-rt}\Psi(S_{t})]\frac{1}{\delta}$ then \begin{align} P(J_{\nu}>\alpha)\leq\frac{1}{\alpha}E[J_{\nu}]\leq\frac{1}{\alpha}E[\sup_{t\in[0,T]}e^{-rt}\Psi(S_{t})]<\delta \end{align} Then \begin{align} E[J_{\nu}1_{J_{\nu}>\alpha}]=\int_{J_{\nu>\alpha}}J_{\nu}dP\leq\int_{J_{\nu}>\alpha}E[sup_{t\in[0,T]}e^{-rt}\Psi(S_{t})|\mathcal F_{\nu}]dP\\ =\int_{J_{\nu}>\alpha}\sup_{t\in[0,T]}e^{-rt}\Psi(S_{t})dP<\epsilon \end{align} I don't get the last estimation $<\epsilon$. Because in the beginning we only assumed, that there exists a set $A$ which doesn't really has something to do with the set $\{J_{\nu}>\alpha\}$. I hope you can help me out. Best regards

Answer 1

Instead of

$$\forall \varepsilon>0 \exists \delta>0, A \in \mathcal{F}_T, \mathbb{P}(A) <\delta: \int_A \sup_{t \in [0,T]} e^{-rt} \Psi(S_t) \, d\mathbb{P}<\varepsilon$$

it should read

$$\forall \varepsilon>0 \exists \delta>0: A \in \mathcal{F}_T, \mathbb{P}(A) <\delta \Rightarrow \int_A \sup_{t \in [0,T]} e^{-rt} \Psi(S_t) \, d\mathbb{P}<\varepsilon.$$

The second statement follows from the following theorem:

Theorem: Let $f \in L^1(\mathbb{P})$ and $\varepsilon>0$. Then there exists $\delta>0$ such that for any $A$ with $\mathbb{P}(A)<\delta$ we have $$\left| \int_A f \, d\mathbb{P} \right| <\varepsilon.$$

Proof: By the Sombrero lemma, we can find a step function $f_n$ of the form $$f_n = \sum_{j=1}^N c_j 1_{A_j}$$ (where $c_j$ are constants and $A_j$ measurable sets) such that $\|f_n-f\|_{L^1} \leq \frac{\varepsilon}{2}$. Since $f_n$ is a step function (hence bounded), we have

$$\int_A |f_n| \, d\mathbb{P} \leq \max_{1 \leq j \leq N} |c_j| \cdot \mathbb{P}(A) < \frac{\varepsilon}{2}$$

for any $A$ such that $$\mathbb{P}(A) < \delta := \frac{\varepsilon}{2 \max_j |c_j|}.$$ Consequently, by the triangle inequality,

$$\begin{align*} \left| \int_A f \, d\mathbb{P} \right| &\leq \int_A |f_n-f| \, d\mathbb{P} + \int_A |f_n| \, d\mathbb{P} \\ &\leq \|f_n-f\|_{L^1} + \int_A |f_n| \, d\mathbb{P} < \varepsilon. \end{align*}$$

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Therefore, the last estimate follows for $A = \{J_{\nu}>\alpha\}$ if we choose $\alpha$ sufficiently large.