Let $L_T$ the local time of a Brownian motion. Roughly speaking, it's the time spend by a BM at $0$, i.e. $$m\{s\in [0,T]\mid B_s=0\}.$$
Now, I know that $$\mathbb P\{L_T\geq x\}=2\mathbb P\{B_T\geq x\}.$$
And since $$2\mathbb P\{B_T\geq T\}>0,$$ there is no chance to have $$\mathbb P\{L_T\leq T\}=1.$$
I think this is rather strange. What am I mistaken here ? How is it possible that $(B_t)_{t\in [0,T]}$ spend more time in $0$ than $T$ ?
Things might be clearer from a wider perspective. The Brownian motion has a local time $L^x_t$ at each level $x\in\Bbb R$ (not just $0$). The collection $(L^x_t)_{x\in\Bbb R, t\ge 0}$ can be chosen to be jointly continuous in $(x,t)$, a.s., and then serves as occupation density: $$ \int_0^T f(B_t)\,dt =\int_{\Bbb R} L^x_T \, f(x)\,dx, $$ for each bounded measurable $f$ and each $T>0$, a.s. In particular, you have $$ \int_{\Bbb R} L^x_T\, dx = T. $$