Wald's Identity on exponential martingale Given a Brownian motion $W$ on the probability space $(\Omega, \mathcal{F}_t^W,P)$, define the process $Z_t = \exp(\mu W_t - \frac{1}{2}\mu t^2)$  and a measure $P^{\mu}(A) = E(1_AZ_t), \ A \in \mathcal{F}_t^W$. Show that, for the stopping time $T$ with $P(T < \infty) = 1$, the Wald identity 
$$E(\exp(\mu W_T - \frac{1}{2}\mu^2T))=1$$ holds iff $$P^\mu((T < \infty)=1$$
To prove the sufficiency, notice that $Z_t$ is indeed a martingale and hence $Z_{t\wedge T}$ is also a martingale, hence
$$E(\exp(\mu W_{t\wedge T} - \frac{1}{2}\mu^2(t\wedge T)))=1$$
How can we make use of the given condition to show 
$$1=E(\lim_{t\to\infty}\exp(\mu W_{t\wedge T} - \frac{1}{2}\mu^2(t\wedge T)))=E(\exp(\mu W_T - \frac{1}{2}\mu^2T))?$$
 A: Since $P^{\mu}$ is a probability measure on $(\Omega,\mathcal{F}_\infty^W)$, it is continuous from below,
$$P^{\mu}(T<\infty)=\lim_{k\to\infty}P^{\mu}(T\leq k)$$
Note that $\{T\leq k\}\in\mathcal{F}_k^W$, so we have $P^{\mu}(T\leq k)=\mathbb{E}[1_{\{T\leq k\}}Z_k]$. Thus,
$$P^{\mu}(T<\infty)=1 \iff \lim_{k\to\infty}\mathbb{E}[1_{\{T\leq k\}}Z_k]=1$$
Notice that $\mathbb{E}[Z_{T\wedge t}]=1$ and $\mathbb{E}[Z_t]=1$ for all $t\geq 0$,
$$1 = \mathbb{E}[Z_{T\wedge t}] = \mathbb{E}[Z_T1_{\{T\leq t\}}] + \mathbb{E}[Z_t1_{\{T>t\}}]$$
$$1 = \mathbb{E}[Z_t] = \mathbb{E}[Z_t1_{\{T\leq t\}}] + \mathbb{E}[Z_t1_{\{T>t\}}]$$
This implies that $\mathbb{E}[Z_T1_{\{T\leq t\}}]=\mathbb{E}[Z_t1_{\{T\leq t\}}]$ for all $t\geq 0$. Therefore,
$$P^{\mu}(T<\infty)=1 \iff \lim_{k\to\infty}\mathbb{E}[1_{\{T\leq k\}}Z_T]=1$$
Notice that $1_{\{T\leq k\}}Z_T$ is nonnegative and increasing in $k$, so by MCT,
$$\lim_{k\to\infty}\mathbb{E}[1_{\{T\leq k\}}Z_T]=\mathbb{E}[1_{\{T<\infty\}}Z_T]$$
Since $P(T<\infty)=1$, we have $\mathbb{E}[1_{\{T<\infty\}}Z_T]=\mathbb{E}[Z_T]$. Thus,
$$P^{\mu}(T<\infty)=1 \iff\mathbb{E}[Z_T]=1$$
