Total Variation of a Density Measure

From a Density to a Variation Measure

A signed density can contribute positive mass on some parts of a set and negative mass on others. The signed measure records the net contribution, so cancellation can make its value small—or zero—even when the density has substantial magnitude. Total variation removes that cancellation. For an integrable real density, it is itself given by a density: the absolute value of the original one.

Let $(X,\mathcal{M},\mu)$ be any measure space; no finiteness or sigma-finiteness assumption on $\mu$ is needed. Let $f:X\to\mathbb{R}$ be measurable and integrable, meaning $\int_X |f|\,d\mu<\infty$. Define the finite signed measure $\nu_f$ on $\mathcal{M}$ by $\nu_f(E)=\int_E f\,d\mu$. The integrals here are finite for every $E\in\mathcal{M}$, since $\int_E|f|\,d\mu\le\int_X|f|\,d\mu$. Countable additivity follows by writing $f=f^+-f^-$ and applying Monotone Convergence to each nonnegative part over disjoint measurable sets.

For a finite signed measure $\nu$, its total variation on $E\in\mathcal{M}$ is the supremum of sums of absolute values over all finite measurable partitions of $E$:

|\nu|(E)=\sup\left\{\sum_{j=1}^{k}|\nu(E_j)|:\ E=\bigcup_{j=1}^{k}E_j,\ E_i\cap E_j=\varnothing\text{ for }i\ne j,\ E_j\in\mathcal{M}\right\}.

Empty pieces are permitted and do not affect the sum. This set function is the total variation measure, as established in the earlier material on total variation. In particular, $|\nu|(E)$ is not generally the number $|\nu(E)|$: the former allows a partition that separates positive and negative contributions.

Theorem: For the measure space and integrable real-valued $f$ above, the total variation measure of $\nu_f(E)=\int_E f\,d\mu$ satisfies $$ |\nu_f|(E)=\int_E |f|\,d\mu\qquad\text{for every }E\in\mathcal{M}. $$ Both sides are finite. The equality is an equality of set functions at every measurable $E$, not merely an almost-everywhere statement.

Proof. Fix $E\in\mathcal{M}$. First consider any finite measurable partition $E=\bigcup_{j=1}^{k}E_j$. For each piece, the integral inequality $\left|\int_{E_j}f\,d\mu\right|\le\int_{E_j}|f|\,d\mu$ follows by integrating $-|f|\le f\le |f|$. The pieces are disjoint, so finite additivity of the nonnegative integral gives $$ \sum_{j=1}^{k}|\nu_f(E_j)| =\sum_{j=1}^{k}\left|\int_{E_j}f\,d\mu\right| \le\sum_{j=1}^{k}\int_{E_j}|f|\,d\mu =\int_E|f|\,d\mu. $$ Taking the supremum over all such partitions proves $|\nu_f|(E)\le\int_E|f|\,d\mu$.

For the reverse inequality, use the particular measurable partition $E=E_+\cup E_-$, where $E_+=E\cap\{f\ge0\}$ and $E_-=E\cap\{f<0\}$. These pieces are disjoint, including when either is empty. On $E_+$, $f=|f|$, and on $E_-$, $f=-|f|$. Therefore $$ |\nu_f(E_+)|+|\nu_f(E_-)| =\int_{E_+}|f|\,d\mu+\int_{E_-}|f|\,d\mu =\int_E|f|\,d\mu. $$ This partition is among those in the supremum defining $|\nu_f|(E)$, so $|\nu_f|(E)\ge\int_E|f|\,d\mu$. Combining the two inequalities proves the formula. $\square$

The proof explains why the formula works: no partition can produce a sum larger than the integral of the magnitude, and the partition by the sign of $f$ achieves that bound. If $f=0$ $\mu$-a.e. on $E$, both sides are zero. If one sign region is empty, the same partition argument still applies without change.

Worked Calculations

Worked Example: A Density That Changes Sign at Zero

Take $X=[-2,2]$ with the Lebesgue sigma-algebra and Lebesgue measure $m$, and let $f(x)=x$. Compute $\nu_f(E)$ and $|\nu_f|(E)$ for $E=[-1,2]$.

The density is integrable because $\int_{-2}^{2}|x|\,dx=4<\infty$. On the negative part $[-1,0)$, $f$ is negative; on $[0,2]$, it is nonnegative. The single point $0$ has measure zero, but the chosen sign partition is exact even with the stated endpoint conventions. The signed measure is

\nu_f(E)=\int_{-1}^{2}x\,dx =\left[\frac{x^2}{2}\right]_{-1}^{2} =2-\frac12 =\frac32.

For total variation, integrate the absolute value separately on the two sign regions: $$ |\nu_f|(E)=\int_{-1}^{0}(-x)\,dx+\int_{0}^{2}x\,dx =\left[-\frac{x^2}{2}\right]_{-1}^{0}+\left[\frac{x^2}{2}\right]_{0}^{2} =\frac12+2=\frac52. $$ Check: the partition $\{[-1,0),[0,2]\}$ gives signed values $-\tfrac12$ and $2$, so the sum of their absolute values is $\tfrac12+2=\tfrac52$. The full-set signed value is only $\tfrac32$, because the negative contribution cancels part of the positive one.

Worked Example: A Piecewise Constant Density

Let $X=[0,3]$ with Lebesgue measure, and define $f=2$ on $[0,1]$ and $f=-3$ on $(1,3]$. The values at the shared boundary do not cause ambiguity: the two specified pieces are disjoint and cover $X$. For $E=[\tfrac12,2]$, compute the signed measure and variation.

The positive portion of $E$ is $[\tfrac12,1]$, of length $1-\tfrac12=\tfrac12$. The negative portion is $(1,2]$, of length $2-1=1$. Consequently, $$ \nu_f(E)=2\left(\frac12\right)-3(1)=1-3=-2, \qquad |\nu_f|(E)=2\left(\frac12\right)+3(1)=1+3=4. $$ The value $-2$ is the net signed contribution; the variation is $4$. Check: the sign partition of $E$ has pieces with signed measures $1$ and $-3$. Their absolute values sum to $1+3=4$, matching the density formula.

For an additional check on the whole space, the positive contribution is $2\cdot1=2$, and the negative contribution has magnitude $3\cdot2=6$. Thus $\nu_f(X)=2-6=-4$, whereas $|\nu_f|(X)=2+6=8$. Directly, $\int_0^1 2\,dx+\int_1^3 3\,dx=2+6=8$.

Worked Example: A Symmetric Set with Zero Signed Measure

Let $X=[0,2]$ with Lebesgue measure and $f(x)=x-1$. For $E=[0,\tfrac12]\cup[\tfrac32,2]$, calculate both quantities. The intervals are disjoint, and their endpoints are included as written.

On $[0,\tfrac12]$, the density is negative, while on $[\tfrac32,2]$ it is positive. The signed contribution of the left interval is $$ \int_0^{1/2}(x-1)\,dx =\left[\frac{x^2}{2}-x\right]_0^{1/2} =\frac18-\frac12 =-\frac38. $$ The signed contribution of the right interval is $$ \int_{3/2}^{2}(x-1)\,dx =\left[\frac{x^2}{2}-x\right]_{3/2}^{2} =0-\left(\frac98-\frac32\right) =\frac38. $$ Hence $\nu_f(E)=-\tfrac38+\tfrac38=0$.

For variation, take absolute values before integrating. On the left, $|x-1|=1-x$, and on the right, $|x-1|=x-1$: $$ |\nu_f|(E)=\int_0^{1/2}(1-x)\,dx+\int_{3/2}^{2}(x-1)\,dx =\left[x-\frac{x^2}{2}\right]_0^{1/2} +\left[\frac{(x-1)^2}{2}\right]_{3/2}^{2} =\frac38+\frac38=\frac34. $$ Check: the two pieces form a sign partition of $E$, with signed measures $-\tfrac38$ and $\tfrac38$. Their absolute values sum to $\tfrac38+\tfrac38=\tfrac34$, even though the signed measure of their union is zero.

Jordan Parts from the Density

The density formula also identifies the positive and negative measures in the Jordan decomposition. Define the positive and negative parts of the real function pointwise by $f^+=\max(f,0)$ and $f^-=\max(-f,0)$. Then $f=f^+-f^-$ and $|f|=f^++f^-$. Both $f^+$ and $f^-$ are measurable, nonnegative, and integrable because they are bounded above by $|f|$.

Corollary: For the finite signed measure $\nu_f$, define $\nu_f^+(E)=\int_E f^+\,d\mu$ and $\nu_f^-(E)=\int_E f^-\,d\mu$. These are finite positive measures, $\nu_f=\nu_f^+-\nu_f^-$, and $$ |\nu_f|=\nu_f^++\nu_f^-. $$ Moreover, $\nu_f^+$ and $\nu_f^-$ are mutually singular.

Proof. Nonnegative integrable densities define finite positive measures by countable additivity of the nonnegative integral. Since $f=f^+-f^-$, integration over any $E\in\mathcal{M}$ gives $\nu_f(E)=\nu_f^+(E)-\nu_f^-(E)$. Since $|f|=f^++f^-$, the theorem gives $|\nu_f|(E)=\int_E|f|\,d\mu=\nu_f^+(E)+\nu_f^-(E)$ for every $E$. To verify mutual singularity, let $P=\{f\ge0\}$ and $N=\{f<0\}$. On $N$, $f^+=0$, so $\nu_f^+(N)=0$; on $P$, $f^-=0$, so $\nu_f^-(P)=0$. The measurable sets $P,N$ are disjoint and cover $X$, which is precisely mutual singularity of these positive measures. Thus these are the Jordan parts. $\square$

A Common Cancellation Error

A frequent mistake is to calculate variation as $|\nu_f(E)|$, rather than integrating $|f|$ over $E$. The third example gives a concrete counterexample: its two signed contributions are $-\tfrac38$ and $\tfrac38$, so $|\nu_f(E)|=|0|=0$. But the partition separating those two signs gives variation $\tfrac38+\tfrac38=\tfrac34$, and indeed $\int_E|f|\,dm=\tfrac34$. The correction is to take the absolute value at the density level before integrating, or equivalently to use the sign partition in the definition of total variation.

On the whole space, the number $|\nu_f|(X)=\int_X|f|\,d\mu$ is sometimes called the total variation norm of the finite signed measure. It is finite by the assumed integrability of $f$. On a subset $E$, however, the same formula is a measure of all positive and negative mass located in $E$, not just the magnitude of the net value $\nu_f(E)$. This distinction is useful whenever a signed density oscillates or changes sign: the signed integral measures net balance, while variation measures total magnitude.

Check Your Understanding

For each question, distinguish the signed integral from the variation integral and use the measurable sign partition when appropriate.

On $[-1,1]$ with Lebesgue measure, let $f(x)=x$. Compute $\nu_f([-1,1])$ and $|\nu_f|([-1,1])$, showing the sign-partition calculation.
Suppose $f\ge0$ $\mu$-a.e. and $f$ is integrable. What does the density formula imply about $|\nu_f|(E)$ and $\nu_f(E)$ for every measurable $E$?
Let $f=1$ on $[0,2]$ and $f=-2$ on $(2,3]$, with Lebesgue measure. Compute the signed measure and variation of $[1,3]$.
Why is the partition into $E\cap\{f\ge0\}$ and $E\cap\{f<0\}$ sufficient to prove the lower bound for total variation?
If $f=0$ $\mu$-a.e. on a measurable set $E$, what are $\nu_f(E)$ and $|\nu_f|(E)$?

Total Variation of a Density Measure

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