Proving that $\{\tau_n=\infty\}=\left\{\int_0^TX_u^2\,du\le n\right\} $

Question

Let $(\Omega,\mathcal A,P)$ probability space, $\mathbb F=\{\mathcal F_t\}_{t\in[0,T]}$ a right-continous and complete filtration (right continuity means that $\mathcal F_t=\bigcap_{\epsilon>0}\mathcal F_{t+\epsilon}$ and completness means that for every $A\in\mathcal A$ with $P(A)=0$, every $\mathcal F_t$ contains every $B$, with $B\subseteq A$).

Let's consider $X\in M^2_{loc}[0,T]$, i.e. $X:\Omega\times[0,T]\to\Bbb R$ is progressively measurable and $\int_0^TX_u^2\,du<+\infty$ a.s.

Define then for every $n\in\Bbb N$ $$ \tau_n:=\inf\left\{t\in[0,T]\;:\;\int_0^tX_u^2\,du>n\right\} $$ with the convention that $\inf\emptyset=+\infty$. So $\tau_n$ is a sequence of $\Bbb F$-stopping times.

I have to prove that $$ \{\tau_n=\infty\}=\left\{\int_0^TX_u^2\,du\le n\right\} $$ Now $\subseteq$ is clear; in order to prove the inverse inclusion I tried to prove that $$ \{\tau_n<\infty\}\subseteq\left\{\int_0^TX_u^2\,du> n\right\} $$ but I have some problems: I argued as follows \begin{align*} \{\tau_n<\infty\} &=\{\tau_n\le T\}=\{\tau_n< T\}\cup \{\tau_n= T\} \end{align*} and clearly $$ \{\tau_n< T\}\subseteq\left\{\int_0^TX_u^2\,du> n\right\} $$ but $$ \{\tau_n= T\}=\left\{\int_0^TX_u^2\,du=n\right\}\;. $$ Thus I proved that $$ \{\tau_n<\infty\}\subseteq\left\{\int_0^TX_u^2\,du\ge n\right\} $$ which is not the inclusion I wanted beacuse of that $\ge$ instead of a $>$.

How can I fix this?

Answer 1

Your mistake is in $\{\tau_n= T\}=\left\{\int_0^TX_u^2\,du=n\right\}\;$. This identity would be correct if the infimum defining $\tau_n$ was not over the compact set $[0,T]$. In the present case we have that actually $\{\tau_n= T\}=\emptyset$. With this fix, the rest of your proof is correct.

Edit on why $\{\tau_n= T\}=\emptyset$:

For $\tau_n=T$ to happen, the set $A=\left\{t\in[0,T]\;:\;\int_0^tX_u^2\,du>n\right\}$ must be non-empty. In particular, $\int_0^TX_u^2\,du>n$, and by continuity, $t\in A$ for some $t