Positive Parts, Negative Parts, and Truncations

Separating the Positive and Negative Values

A measurable function can change sign, be unbounded, or even take the values $+\infty$ and $-\infty$. Two operations isolate its nonnegative and nonpositive contributions: the positive part records the amount above zero, and the negative part records the magnitude below zero. Bounded truncations then clip values outside a chosen interval. These operations are useful throughout measure theory, including when signed functions are later integrated, but measurability alone does not imply integrability.

Unless an example specifies otherwise, let $(X,\mathcal{M},\mu)$ be any measure space; no finiteness, completeness, or $\sigma$-finiteness assumption is imposed. Consider an extended-real-valued measurable function $f:X\to[-\infty,\infty]$. Here measurability means that $\{x:f(x)>a\}\in\mathcal{M}$ for every real $a$. Equalities between functions in this tutorial are pointwise, not merely $\mu$-a.e.

Definition: The positive part and negative part of $f$ are the extended-real-valued functions $$ f^+(x)=\max\{f(x),0\},\qquad f^-(x)=\max\{-f(x),0\}. $$ Both take values in $[0,\infty]$. In particular, $f^+=+\infty$ when $f=+\infty$, and $f^-=+\infty$ when $f=-\infty$.

The negative part is a magnitude, not a signed negative value. Thus $f^-$ is nonnegative, even where $f$ is negative. This convention gives the reconstruction identities $f=f^+-f^-$ and $|f|=f^++f^-$; the first subtraction is always defined because at most one of $f^+$ and $f^-$ can be infinite at a point.

Measurability and Reconstruction

Proposition: On any measure space $(X,\mathcal{M},\mu)$, if $f:X\to[-\infty,\infty]$ is measurable, then $f^+$ and $f^-$ are measurable. Pointwise on $X$, $$ f=f^+-f^-,\qquad |f|=f^++f^-. $$

Proof. For any real $a<0$, the set $\{f^+>a\}$ is $X$, since $f^+\ge0$. For real $a\ge0$, the definition of maximum gives $\{f^+>a\}=\{f>a\}$. These sets are measurable, so $f^+$ is measurable. Likewise, for $a<0$, $\{f^->a\}=X$, while for $a\ge0$, $$ \{f^->a\}=\{-f>a\}=\{f<-a\}. $$ The last set is measurable: its complement is $\{f\ge-a\}$, which is measurable by the level-set criterion for extended-real-measurable functions. Thus $f^-$ is measurable.

To verify the identities, fix $x\in X$. If $f(x)>0$, then $f^+(x)=f(x)$ and $f^-(x)=0$, so their difference is $f(x)$ and their sum is $|f(x)|$. If $f(x)<0$, including $f(x)=-\infty$, then $f^+(x)=0$ and $f^-(x)=-f(x)$, giving the same identities. If $f(x)=0$, both parts are zero. The case $f(x)=+\infty$ is included in the positive case: the parts are $+\infty$ and $0$. Hence both identities hold everywhere. $\square$

Worked Example: A Sign-Changing Function on a Bounded Interval

On $X=[-3,3]$ with its Borel $\sigma$-algebra and Lebesgue measure restricted to $X$, take $f(x)=x-1$. Since this is a continuous real-valued function, it is measurable. Its sign changes at $x=1$, so the definitions give $$ f^+(x)= \begin{cases} 0,&-3\le x\le1,\\ x-1,&1<x\le3, \end{cases} \qquad f^-(x)= \begin{cases} 1-x,&-3\le x<1,\\ 0,&1\le x\le3. \end{cases} $$ For the symmetric truncation with $M=1$, the values are clipped to $[-1,1]$: $$ T_1f(x)= \begin{cases} -1,&-3\le x<0,\\ x-1,&0\le x\le2,\\ 1,&2<x\le3. \end{cases} $$ At $x=-2$, $f=-3$, $f^+=0$, $f^-=3$, and $T_1f=-1$; indeed $0-3=-3$. At $x=1$, $f=0$ and both parts and the truncation are $0$. At $x=3$, $f=2$, $f^+=2$, $f^-=0$, and $T_1f=1$; again the parts recover $f$, while the truncation lies in $[-1,1]$. These substitutions check the formulas on the negative, zero, and positive regions.

Symmetric Bounded Truncation

Definition: For $M>0$, the symmetric truncation of an extended-real-valued measurable function $f$ is $$ T_M(f)=\max\{-M,\min\{f,M\}\}. $$ It clips values below $-M$ to $-M$, values above $M$ to $M$, and leaves values in $[-M,M]$ unchanged.

Proposition: If $f:X\to[-\infty,\infty]$ is measurable on $(X,\mathcal{M},\mu)$, then $T_M(f)$ is measurable and real-valued, and $-M\le T_M(f)\le M$ everywhere. For each $x$, $T_n(f)(x)\to f(x)$ as the positive integer $n\to\infty$, with convergence understood in the extended-real sense.

Proof. Fix $M>0$. The definition gives $T_M(f)(x)=-M$ when $f(x)<-M$, $T_M(f)(x)=f(x)$ when $-M\le f(x)\le M$, and $T_M(f)(x)=M$ when $f(x)>M$. This proves the asserted range and the clipping description, including the endpoint values. To prove measurability, test strict upper level sets. If $a<-M$, then $\{T_M(f)>a\}=X$. If $-M\le a<M$, then $\{T_M(f)>a\}=\{f>a\}$. If $a\ge M$, then $\{T_M(f)>a\}=\varnothing$. All three sets are measurable, so $T_M(f)$ is measurable.

For the limit, fix $x$. If $f(x)$ is finite, every integer $n\ge |f(x)|$ gives $T_n(f)(x)=f(x)$. If $f(x)=+\infty$, then $T_n(f)(x)=n$ for every $n$, which tends to $+\infty$. If $f(x)=-\infty$, then $T_n(f)(x)=-n$, which tends to $-\infty$. These cases establish the pointwise extended-real limit. $\square$

For a nonnegative measurable function $g:X\to[0,\infty]$, the one-sided truncation $g\wedge M=\min\{g,M\}$ is also measurable and takes values in $[0,M]$. It follows either from the level sets $\{g\wedge M>a\}$, or from $g\wedge M=T_M(g)$ when $g\ge0$. This one-sided form is often convenient when only nonnegative values are under consideration.

Worked Example: Truncating an Unbounded Sign-Changing Function

On $\mathbb{R}$ with the Borel $\sigma$-algebra and Lebesgue measure, define $f(0)=0$ and $f(x)=1/x$ for $x\ne0$. This function is measurable: it is continuous on each of the measurable sets $(-\infty,0)$ and $(0,\infty)$, and its value on the remaining singleton is specified separately. Its positive and negative parts are $$ f^+(x)= \begin{cases}1/x,&x>0,\\0,&x\le0,\end{cases} \qquad f^-(x)= \begin{cases}-1/x,&x<0,\\0,&x\ge0.\end{cases} $$ For $M=2$, if $0<x<1/2$, then $1/x>2$, so $T_2(f)(x)=2$. If $x\ge1/2$, then $0<1/x\le2$, so $T_2(f)(x)=1/x$. On the negative side, $-1/2<x<0$ implies $1/x<-2$, so $T_2(f)(x)=-2$; when $x\le-1/2$, $-2\le1/x<0$, so $T_2(f)(x)=1/x$. At zero the truncation is zero. Thus $$ T_2(f)(x)= \begin{cases} -2,&-1/2<x<0,\\ 1/x,&x\le-1/2,\\ 0,&x=0,\\ 1/x,&x\ge1/2,\\ 2,&0<x<1/2. \end{cases} $$ At $x=1/4$, $f=4$, $f^+=4$, $f^-=0$, and $T_2(f)=2$. At $x=-1/4$, $f=-4$, $f^+=0$, $f^-=4$, and $T_2(f)=-2$. At $x=1$ and $x=-1$, the function values are $1$ and $-1$, respectively, so truncation leaves both unchanged. These values verify the signs and the clipping threshold directly.

Worked Example: Parts and Truncation with Infinite Values

On $\mathbb{R}$ with its Borel $\sigma$-algebra and Lebesgue measure, let $A=[0,1)$, $B=[1,2)$, $C=[2,\infty)$, and $D=(-\infty,0)$. These sets are disjoint and cover $\mathbb{R}$, with endpoints assigned as displayed. Define $f=+\infty$ on $A$, $f=-3$ on $B$, $f=-\infty$ on $C$, and $f=0$ on $D$. The function is measurable because each of its level sets is formed from these Borel sets. Its parts are $$ f^+= \begin{cases}+\infty,&x\in A,\\0,&x\in B\cup C\cup D,\end{cases} \qquad f^-= \begin{cases}3,&x\in B,\\+\infty,&x\in C,\\0,&x\in A\cup D.\end{cases} $$ For $M=2$, $T_2(f)$ is $2$ on $A$, $-2$ on $B\cup C$, and $0$ on $D$. At a point in $A$, the reconstructed value is $+\infty-0=+\infty$; at a point in $B$, it is $0-3=-3$; and at a point in $C$, it is $0-(+\infty)=-\infty$. On $D$, both parts vanish. This checks the reconstruction even at both infinite values, while the truncation remains finite everywhere.

Common Mistake and Scope of the Results

A frequent sign error is to define the negative part as $\min\{f,0\}$ and then use it as a nonnegative magnitude. For example, if $f(x)=-3$, this incorrect expression equals $-3$, and subtracting it from $f(x)$ gives $-3-(-3)=0$, not $-3$. The correct negative part is $f^-(x)=\max\{-f(x),0\}=3$, so $f(x)=f^+(x)-f^-(x)=0-3=-3$. Equivalently, the signed quantity $\min\{f,0\}$ is $-f^-$, not $f^-$.

Measurability is preserved by these operations, but further properties require their own hypotheses. In particular, the fact that $T_M(f)$ is bounded and measurable does not by itself assert that $f$ is integrable, nor does it establish an equality between an integral and a limit of integrals. Those are separate questions about the measure space and the integral. The pointwise limit $T_n(f)\to f$ proved above is a statement about values at each point; it is not a claim of uniform convergence or convergence in $L^p(\mu)$.

Check Your Understanding

Use the definitions and check endpoints carefully; all requested equalities are pointwise.

For $f(x)=2-x$ on $[0,4]$, write $f^+$ and $f^-$ piecewise, and verify their difference at $x=0,2,4$.
For an extended-real-measurable $f$, describe $\{f^+>a\}$ for $a<0$ and $a\ge0$. Why do these sets prove measurability?
Compute $T_3(f)$ at a point where $f=-5$, a point where $f=2$, and a point where $f=+\infty$.
Explain why $f^+-f^-$ is defined at every point even when $f$ takes infinite values.
For $f(x)=1/x$ when $x\ne0$, with $f(0)=0$, find the set of $x>0$ where $T_1(f)(x)=1$, including whether the endpoint belongs.

Positive Parts, Negative Parts, and Truncations

What You'll Learn