Absolute Values and Integrability

What Absolute Values Tell Us About Integrability

The previous tutorial, Additivity Over Intervals, showed how integrals over subintervals combine. A different question is whether changing a function by taking its absolute value preserves integrability. The Absolute-Value Inequality for the Riemann Integral theorem, established in Properties of the Riemann Integral, already answers the forward question: if $f$ is Riemann integrable, then $|f|$ is Riemann integrable. It also gives an inequality between their integrals. We will use that result rather than prove it again.

Taking an absolute value removes the sign of each function value, but the resulting function can still encode useful information. In particular, it separates $f$ into a nonnegative part and a part recording the magnitude of its negative values. Those parts lead to integral identities and provide formulas for comparing two integrable functions pointwise.

Definition: For a function $f:[a,b]\to\mathbb{R}$, define its positive part and negative part by $$ f^+(x)=\max\{f(x),0\}, \qquad f^-(x)=\max\{-f(x),0\}. $$ Both functions are nonnegative. The negative part $f^-$ records the magnitude of the negative values of $f$, rather than retaining their negative sign.

The definitions can be rewritten using absolute values. For a real number $y$, if $y\geq0$, then $(|y|+y)/2=y$ and $(|y|-y)/2=0$. If $y<0$, then $(|y|+y)/2=0$ and $(|y|-y)/2=-y$. Consequently, for every $x\in[a,b]$,

f^+(x)=\frac{|f(x)|+f(x)}{2}, \qquad f^-(x)=\frac{|f(x)|-f(x)}{2}.

These formulas make it possible to apply earlier integrability results directly. They also give two identities that hold pointwise:

f=f^+-f^-, \qquad |f|=f^++f^-.

Positive and Negative Parts of an Integrable Function

The next theorem packages these identities together with the integral. Its proof uses the Absolute-Value Inequality for the Riemann Integral theorem to obtain integrability of $|f|$, and then uses linearity. No separate integrability argument for the positive and negative parts is needed.

Theorem (Positive-Negative Decomposition): If $f:[a,b]\to\mathbb{R}$ is Riemann integrable, then $f^+$ and $f^-$ are Riemann integrable, and $$ \int_a^b f(x)\,dx = \int_a^b f^+(x)\,dx-\int_a^b f^-(x)\,dx, \qquad \int_a^b |f(x)|\,dx = \int_a^b f^+(x)\,dx+\int_a^b f^-(x)\,dx. $$

Proof. Since $f$ is Riemann integrable, the Absolute-Value Inequality for the Riemann Integral theorem implies that $|f|$ is Riemann integrable. The function $f$ is integrable by hypothesis, so linearity of the Riemann integral shows that $(|f|+f)/2$ and $(|f|-f)/2$ are integrable. These functions are $f^+$ and $f^-$, respectively, so both parts are integrable.

The pointwise identities $f=f^+-f^-$ and $|f|=f^++f^-$, together with linearity, now give

\begin{aligned} \int_a^b f(x)\,dx &=\int_a^b f^+(x)\,dx-\int_a^b f^-(x)\,dx,\\ \int_a^b |f(x)|\,dx &=\int_a^b f^+(x)\,dx+\int_a^b f^-(x)\,dx. \end{aligned}

This proves both integrability and the asserted integral identities. $\square$

The two identities distinguish signed accumulation from total magnitude. The integral of $f$ subtracts the contribution from its negative values, while the integral of $|f|$ adds that magnitude instead. The identities do not assert that either part is zero when $f$ changes sign.

Worked Example: Separating Positive and Negative Contributions

Define $f:[0,3]\to\mathbb{R}$ by $f(x)=2$ for $0\leq x<1$, $f(x)=-1$ for $1\leq x<2$, and $f(x)=3$ for $2\leq x\leq3$. This step function is Riemann integrable. Apart from endpoint values, its positive part equals $2$ on an interval of length $1$ and $3$ on an interval of length $1$; its negative part equals $1$ on an interval of length $1$. Finite point changes do not affect the integral, so

\int_0^3 f^+(x)\,dx=2(1)+3(1)=5, \qquad \int_0^3 f^-(x)\,dx=1(1)=1.

The Positive-Negative Decomposition theorem therefore gives

\int_0^3 f(x)\,dx=5-1=4, \qquad \int_0^3 |f(x)|\,dx=5+1=6.

Directly adding the signed contributions gives $2-1+3=4$, while adding their magnitudes gives $2+1+3=6$. The different answers reflect the different roles of $f$ and $|f|$.

Maxima and Minima of Integrable Functions

Absolute values also give formulas for the pointwise maximum and minimum of two functions. For real numbers $u$ and $v$, the larger value is $(u+v+|u-v|)/2$, and the smaller is $(u+v-|u-v|)/2$. To check the first formula, if $u\geq v$, then $|u-v|=u-v$, so the expression is $u$; if $u<v$, then $|u-v|=v-u$, so it is $v$. The second formula gives $v$ in the first case and $u$ in the second.

Theorem (Integrability of Maxima and Minima): If $f,g:[a,b]\to\mathbb{R}$ are Riemann integrable, then the functions $$ \max\{f,g\} \quad\text{and}\quad \min\{f,g\}, $$ defined pointwise, are Riemann integrable.

Proof. Linearity of the Riemann integral implies that $f-g$ is Riemann integrable. By the Absolute-Value Inequality for the Riemann Integral theorem, $|f-g|$ is integrable. Applying the pointwise real-number formulas at each $x$ gives

\max\{f(x),g(x)\} = \frac{f(x)+g(x)+|f(x)-g(x)|}{2},

\min\{f(x),g(x)\} = \frac{f(x)+g(x)-|f(x)-g(x)|}{2}.

Each right-hand side is a linear combination of integrable functions. Linearity therefore implies that both the maximum and the minimum are Riemann integrable. $\square$

The theorem is useful when a function is described by whichever of two formulas is larger or smaller at each point. It avoids having to locate every crossing point before deciding whether the resulting function is integrable. The formulas prove integrability even if the functions cross many times.

Worked Example: The Maximum of Two Continuous Functions

Let $f(x)=x^2$ and $g(x)=2-x$ on $[0,2]$. Both are continuous and therefore Riemann integrable. The maximum function is

h(x)=\max\{x^2,2-x\} =\frac{x^2+2-x+|x^2+x-2|}{2}.

The integrability theorem shows that $h$ is Riemann integrable. The two expressions agree where $x^2=2-x$, or $x^2+x-2=0$. Factoring gives $(x+2)(x-1)=0$, and the only solution in $[0,2]$ is $x=1$. For $0\leq x<1$, $(x+2)(x-1)<0$, so $x^2<2-x$. For $1<x\leq2$, the product is positive, so $x^2>2-x$. Thus the same maximum can also be written piecewise as

h(x)= \begin{cases} 2-x,&0\leq x\leq1,\\ x^2,&1\leq x\leq2. \end{cases}

The two formulas agree at $x=1$, since $2-1=1^2=1$. The pointwise absolute-value formula proves integrability without relying on the piecewise description; the crossing calculation simply identifies which function supplies the maximum on each part.

Absolute-Value Integrability Does Not Work Backwards

The implication from integrability of $f$ to integrability of $|f|$ cannot generally be reversed. Taking absolute values can erase oscillation between positive and negative values. The following example makes that loss of information exact: the absolute value is constant even though the original function is not Riemann integrable.

Worked Example: An Integrable Absolute Value with a Nonintegrable Function

On $[0,1]$, define $f(x)=1$ when $x$ is rational and $f(x)=-1$ when $x$ is irrational. At every point, $|f(x)|=1$, so $|f|$ is the constant function $1$, which is Riemann integrable and has integral $1$.

To see why $f$ itself is not Riemann integrable, consider any partition of $[0,1]$. Every subinterval of positive length contains both a rational number and an irrational number. Thus on each such subinterval the supremum of $f$ is $1$ and the infimum is $-1$. If the subinterval lengths are $\Delta x_1,\ldots,\Delta x_n$, its upper sum is $\sum_{i=1}^n\Delta x_i=1$, and its lower sum is $-\sum_{i=1}^n\Delta x_i=-1$. Therefore the upper sum minus the lower sum is $2$ for every partition. The Darboux Criterion requires partitions with arbitrarily small upper-minus-lower sum for an integrable function, so $f$ is not Riemann integrable.

This example shows why integrability of $|f|$ alone is not enough: the absolute value does not record how rapidly, or even whether, the signs of $f$ vary.

Using the Results Carefully

When $f$ is integrable, the Absolute-Value Inequality for the Riemann Integral theorem supplies integrability of $|f|$, and the Positive-Negative Decomposition theorem then supplies integrability of $f^+$ and $f^-$. For two integrable functions, applying the same absolute-value result to their difference gives integrability of their pointwise maximum and minimum. These are forward implications with a specific hypothesis: the original function or functions must be integrable.

The counterexample also warns against confusing a pointwise operation with an equivalence of integrability. The map $f\mapsto |f|$ discards signs, and so it may remove discontinuous oscillation. By contrast, the formulas for maxima and minima use $|f-g|$ together with the integrable functions $f$ and $g$; this is why the theorem about maxima and minima has enough information to prove its conclusion.

In applications, it is helpful to write the relevant pointwise identity before making an integrability claim. For positive and negative parts, use $(|f|+f)/2$ and $(|f|-f)/2$. For a maximum or minimum, apply the corresponding formula to $f-g$. Then verify that each function on the right is integrable using the earlier theorems on absolute values and linearity. This keeps the direction of every implication explicit.

Start with the integrability hypothesis.
Check that $f$, or both $f$ and $g$, are Riemann integrable on the interval.

Form the absolute value you need.
Use the Absolute-Value Inequality for the Riemann Integral theorem on $f$ or on $f-g$.

Apply a pointwise identity.
Express positive and negative parts, a maximum, or a minimum using sums and absolute values.

Use linearity and keep the direction clear.
These arguments prove integrability of the transformed function from integrability of the original function or functions; the converse need not hold.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

Write $f^+$ and $f^-$ in terms of $f$ and $|f|$, and state how their difference and sum relate to $f$ and $|f|$.
If $f$ is Riemann integrable, what are the integral formulas for $f$ and $|f|$ in terms of $f^+$ and $f^-$?
Explain why the maximum of two Riemann integrable functions is Riemann integrable.
For the function equal to $1$ on rationals and $-1$ on irrationals in $[0,1]$, why is its absolute value integrable while the function itself is not?
Does integrability of $|f|$ imply integrability of $f$ in general? Give the example from this tutorial that justifies your answer.