the unbounded variation of Brownian paths
The entire reason stochastic integration needs to be reinvented from scratch is hidden in a single pathological fact about Brownian sample paths: they have infinite total variation on every interval. If you wanted to define an integral of the form integral H_s dB_s the way you would define integral H_s df_s for an ordinary function f, you would reach for the Riemann-Stieltjes integral, whose convergence is guaranteed precisely when the integrator f has bounded variation. Brownian motion fails that hypothesis completely, so the classical pathwise construction simply does not exist, and a new probabilistic idea (the Ito integral, built on the L^2 isometry) is forced upon us.
Total variation of a function f over [0, t] is the supremum, over all partitions 0 = t_0 < t_1 < ... < t_n = t, of the sum of |f(t_k) - f(t_{k-1})|. A function of bounded variation is one for which this supremum is finite; equivalently, it is the difference of two increasing functions, and Riemann-Stieltjes (or, in the Lebesgue picture, the associated signed measure) makes sense against it. For Brownian motion the picture is the opposite: as the partition mesh shrinks, the sum of absolute increments grows without bound almost surely, so the total variation of t -> B_t is infinite on every interval, no matter how short. The deep companion fact is that while the first-order sums blow up, the SECOND-order sums (the sum of squared increments) converge to a finite, deterministic limit: the quadratic variation [B]_t = t. Brownian motion is, intuitively, exactly 'half as rough' as it would need to be to have bounded variation, and exactly 'rough enough' to have a non-trivial quadratic variation.
Why this matters: the unbounded variation is not a technicality, it is the source of the entire second-order calculus. Because integral H dB cannot be a pathwise Riemann-Stieltjes integral, the choice of evaluation point in the Riemann sum (left endpoint versus midpoint) genuinely changes the answer, which is why the Ito integral (left endpoint) and the Stratonovich integral (midpoint) differ. And because (dB)^2 behaves like dt rather than being negligible, Ito's formula carries its famous second-order correction term. The honest caveat: a continuous process can have unbounded variation yet still be perfectly tractable; conversely, any continuous process of bounded variation has zero quadratic variation, so it contributes nothing to the Ito correction. Bounded variation and non-trivial quadratic variation are mutually exclusive for continuous paths.
Partition [0, 1] into n equal pieces. The sum of absolute Brownian increments, sum |B_{k/n} - B_{(k-1)/n}|, has expectation about sqrt(2/pi) * sqrt(n), which goes to infinity as n grows — first-order variation diverges. But the sum of SQUARED increments, sum (B_{k/n} - B_{(k-1)/n})^2, has expectation exactly 1 and variance of order 1/n, so it converges to 1 = [B]_1 — second-order variation is finite.
First-order sums diverge (no bounded variation, so no Riemann-Stieltjes); second-order sums converge to t (the quadratic variation that drives Ito's formula).
Almost every Brownian path is also nowhere differentiable, which is the same roughness seen pointwise; do not try to write dB/dt — white noise is a distribution, not a function, and integral H dB must be defined as a whole, not as integral H (dB/dt) dt.