Let $(B_t)_{t\geq 0}$ be a Brownian motion. Define $X_t=tB_{1/t}$ if $t>0$ and $X_t=0$ if $t=0$.
In proving that $X$ is a Brownian motion, A. Klenke in Theorem 21.14 of his book Probability Theory (3rd version) checks for the path continuity that this term is bounded by zero:
$$ \limsup_{t\downarrow0} X_t = \limsup_{t\to\infty} \frac{1}{t}B_t\leq\limsup_{n\to\infty}\frac{1}{n}B_n+\limsup_{n\to\infty}\frac{1}{n}\sup\{B_t-B_n,t\in[n,n+1]\} $$
He uses a SLLN argument for the first term in the RHS and a Borel Cantelli one for the second term to show that they tend to zero: this part is clear to me.
He then concludes that $X$ is continuous at 0. Why is sufficient to check only the $\limsup_{t\downarrow0}X_t$?
I would have checked that also this is bounded by zero with similar arguments:
$$ \liminf_{t\downarrow0} X_t = \limsup_{t\to\infty} \frac{1}{t}B_t\geq\liminf_{n\to\infty}\frac{1}{n}B_n-\liminf_{n\to\infty}\frac{1}{n}\sup\{B_n-B_t,t\in[n,n+1]\} $$
Why could we skip the "$\liminf$" argument?
Please, let me know if something is not clear. Thanks for the help.