Covariance of overlapping Brownian increments

Bilinearity and the Brownian covariance kernel give $\mathrm{Cov}(B_t-B_s,B_v-B_u)=\min (t,v)-\min (t,u)-\min (s,v)+\min (s,u)\mathrm{.}$ Every term is finite because the Brownian values are square-integrable.

Both sides of the claimed formula are unchanged when the ordered pairs $(s,t)$ and $(u,v)$ are exchanged, the left side by symmetry of covariance. It therefore suffices to assume $s\le u$. If $t\le u$, the four minima in step 1.1 are respectively $t,t,s,s$, so the covariance is zero; also $\min (t,v)\le u=\max (s,u)$, so the stated overlap length is zero. This includes $t=u$ and all zero-length first intervals.

Still assuming $s\le u$, suppose $u<t\le v$. The four minima in step 1.1 are $t,u,s,s$, so the covariance is $t-u$. Here $\min (t,v)=t$ and $\max (s,u)=u$, giving the same positive overlap length. This case includes $s=u$ and $t=v$, but excludes $t=u$, already handled in step 2.1.

Finally, if $s\le u\le v<t$, the four minima in step 1.1 are $v,u,s,s$, so the covariance is $v-u$. Here $\min (t,v)=v$ and $\max (s,u)=u$. The value is zero exactly when $u=v$, so zero-length second intervals are included. These three cases exhaust $s\le u$; pair symmetry handles $u<s$.

Steps 2.1, 3.1, and 4.1 prove the formula for every allowed endpoint order, including coincident endpoints, $s=t$, $u=v$, and $s=u=t=v=0$. There is no empty family or biconditional. AC is used only through [F1]; expanding four covariances and comparing endpoints uses no further choice.