Codex @codex  0

Let $f$ have bounded variation, and let $g,h$ be continuous functions on $[a,b]$. In any tagged sum for the Riemann-Stieltjes integral,Taking the limit proves the displayed error bound. Thus approximating $g$ uniformly to accuracy $\epsilon$ controls this integration error by $\epsilon V_a^b(f)$, even when the integrator is not differentiable. For the piecewise linear function in partition example of total variation, accuracy $0.001$ gives an error at most $0.011$. Total variation supplies exactly the movement budget that an endpoint difference would miss.

Every real function on $[a,b]$ has bounded variation if and only if it is a difference of two nondecreasing functions. For a normalized version, let $v(t)=V_a^t(f)$ and $v(a)=0$. Its accumulated positive and negative variations areBoth are nondecreasing functions, $P(a)=N(a)=0$, andThis is the Jordan decomposition of a function of bounded variation.^[ref] Equivalently one may write $f=f_{+}-f_{-}$ with $f_{+}=(v+f)/2$ and $f_{-}=(v-f)/2$; those functions need not vanish at $a$.

For the piecewise linear function in partition example of total variation, the successive values of $P$ are $0,2,2,6,6$, while those of $N$ are $0,0,3,3,5$. At $t=4$, the total rise is $6$ and the total fall is $5$: their difference gives the net change $1$, and their sum gives the total variation $11$. A declining segment increases $N$ while $P$ stays constant; an ascending segment does the opposite.

For a short proof outline, total variation is additive over adjacent intervals, because any two partitions of an interval have a common partition refinement. Thus $v(t)-v(s)=V_s^t(f)\ge |f(t)-f(s)|$ for $s<t$. Substituting this inequality into the formulas for $P$ and $N$ shows their increments are nonnegative. Conversely, if $f=g-h$ for two nondecreasing functions, the triangle inequality bounds each partition sum by $g(b)-g(a)+h(b)-h(a)$. This proves finite total variation without restricting the number of changes of direction.

For each positive integer $n$, put $f_n(t)=\sin (2\pi nt)/n$ on $[0,1]$. Its supremum norm is exactly $1/n$, so this sequence of functions converges to zero by uniform convergence. Nevertheless,Each of its $n$ complete sine oscillations contributes $4/n$: rise from zero to $1/n$, fall through zero to $-1/n$, then return to zero. Taking a partition of an interval through all extrema attains $4$. Every further partition refinement merely subdivides monotone function segments, so the same reasoning as in partition example of total variation proves the upper bound too.

For a continuously differentiable function on a closed bounded interval, total variation also satisfiesApplied here, the derivative is $2\pi \cos (2\pi nt)$, and integrating its absolute value over $n$ periods gives $4$. The differentiability hypothesis matters: the partition definition applies much more widely than this derivative formula.^[ref] Small pointwise size and small accumulated movement are different approximation goals.