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For a bounded real function, inserting partition points increases its lower Darboux sum and decreases its upper Darboux sum. This follows interval by interval because infima increase on smaller sets while suprema decrease, and the new widths sum to the original width. A common partition refinement consequently proves $s(f,D_1)\le S(f,D_2)$ for arbitrary two partitions.

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.

Define a piecewise linear function by joining the five points $(0,0),(1,2),(2,-1),(3,3),(4,1)$. Its net change is $1$, but the increments at these knots are $2,-3,4,-2$. Their absolute values giveThis is its exact total variation, not merely one lower bound. Given any partition of an interval, insert all five knots. The triangle inequality ensures that this partition refinement cannot decrease the sum. On each resulting monotone function segment, the absolute increments telescope, so every refined sum is $11$. The knot partition of an interval itself attains $11$.

Sampling only at $0,4$ gives $1$. Sampling at $0,1,3,4$ gives $2+1+2=5$. Inserting the remaining knot at $2$ reveals the descent and rise that were hidden inside $[1,3]$, giving $11$. These are nested partitions of an interval: their sums increase because each is a partition refinement of the preceding one. Arbitrarily chosen unrelated partitions of an interval need not produce an increasing sequence of sums.

A Riemann integrable function on a closed interval is bounded and has equal upper and lower Darboux integrals. More explicitly, for a partition of an interval $P:a=x_0<x_1<\cdots <x_m=b$, define the Darboux sumsThen Riemann integrability meansand this common value is the Riemann integral. Equivalently, the Riemann integrability criterion requires that for every $\epsilon >0$ some partition of an interval satisfies $U(f,P)-L(f,P)<\epsilon$. Indeed, the common partition refinement of two partitions increases each lower Darboux sum and decreases each upper Darboux sum, so equality of the two extremal values is equivalent to arbitrarily small gaps.

For a partition $D=\{a=x_0<x_1<\cdots <x_n=b\}$, put $\Delta x_k=x_k-x_{k-1}$ andBoundedness makes these finite. The lower Darboux sum and upper Darboux sum areSince $m_k\le M_k$ and the widths are positive, $s(f,D)\le S(f,D)$.

Splitting a partition interval into smaller intervals can only increase each infimum and decrease each supremum. The smaller widths sum to the original width. Its new lower contribution is therefore at least its old lower contribution, and its new upper contribution at most its old upper contribution. Repeating this for every inserted point proves Darboux sum refinement monotonicity:For arbitrary partitions, take their common partition refinement $D_{\ast }=D_1\cup D_2$. ThenNo compatibility of the original partitions is required.

Finally suppose $f\ge 0$ is Riemann integrable. Then $t_k=m_k\Delta x_k\ge 0$, so every product factor is nonnegative and the exponential inequality can be multiplied safely:A lower Darboux sum is no greater than the Riemann integral, since the integral equals the supremum of all lower sums. Monotonicity of the exponential therefore gives the exponential bound for a lower-sum productFor a degenerate interval $a=b$, the empty product and the exponential are both one.

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.