  We give a simple proof that for a continuous local martingale $M_{t}$ $$
\liminf_{\varepsilon\downarrow0}\varepsilon \log Ee^{(1-\varepsilon) \langle
M\rangle_{\infty}/2}<\infty \Longrightarrow E\exp(M_{\infty}-\langle
  M\rangle_{\infty}/2)=1. $$
