Why the Integral Triangle Inequality Matters
The integral combines contributions from different parts of a space. Those contributions can reinforce one another or cancel, so the magnitude of the total integral need not equal the total magnitude of the contributions. The integral triangle inequality gives the fundamental bound: cancellation can reduce the result, but the magnitude of the result cannot exceed the integral of the pointwise magnitude. A direct consequence controls how much an integral can change when its integrand changes.
Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space; no finiteness, sigma-finiteness, or completeness assumption is required. Functions may be real-valued or complex-valued, and all integrals in the statements below are finite Lebesgue integrals. A measurable function \(f\) is integrable when \(\int |f|\,d\mu<\infty\). For a complex-valued measurable function, measurability means that its real and imaginary parts are measurable, and its integral is defined by integrating those parts. We use the earlier results that the integral is linear on integrable functions, that it is monotone for nonnegative measurable functions, and that \(\operatorname{Re}\int f\,d\mu=\int\operatorname{Re}f\,d\mu\).
The Integral Triangle Inequality
Proof. Since \(f\) is integrable, its integral \(z=\int f\,d\mu\) is finite. If \(z=0\), then \(\left|\int f\,d\mu\right|=0\le\int|f|\,d\mu\), because \(|f|\ge0\). This includes the case \(\int|f|\,d\mu=0\). Now suppose \(z\ne0\), and set \(\alpha=\overline{z}/|z|\). Then \(|\alpha|=1\) and \(\alpha z=|z|\). For every \(x\in X\), the elementary complex-number inequality \(\operatorname{Re}w\le |w|\) gives $$ \operatorname{Re}(\alpha f(x))\le|\alpha f(x)|=|f(x)|. $$ Both functions in this comparison are integrable: \(|f|\) is integrable by assumption, and \(|\operatorname{Re}(\alpha f)|\le|f|\). Monotonicity of the integral therefore gives $$ \left|\int f\,d\mu\right| =\operatorname{Re}(\alpha z) =\int\operatorname{Re}(\alpha f)\,d\mu \le\int|f|\,d\mu. $$ Here the equality involving the integral follows from linearity and compatibility of the integral with real parts. This proves the claim for complex-valued \(f\); real-valued functions are included by viewing them as complex-valued functions with zero imaginary part. \(\square\)
The proof uses a phase \(\alpha\) chosen from the integral itself, not from each value \(f(x)\). It rotates the total integral to the nonnegative real axis. At each point, the real part after that rotation is no larger than the pointwise magnitude. Integrating that pointwise comparison yields the result.
Worked Example: Cancellation for a Real-Valued Function
Let \(X=[0,3]\), let \(\mathcal{M}\) be the Lebesgue sigma-algebra restricted to \(X\), and let \(\mu\) be Lebesgue measure. Define \(f=2\) on \([0,1]\) and \(f=-1\) on \((1,3]\). Compute both sides of the inequality.
The two pieces have measures \(1\) and \(2\), respectively. Thus $$ \int_X f\,dm=2\cdot 1+(-1)\cdot 2=2-2=0. $$ On the same pieces, \(|f|\) has values \(2\) and \(1\), so $$ \int_X|f|\,dm=2\cdot1+1\cdot2=2+2=4. $$ Therefore \(\left|\int_X f\,dm\right|=|0|=0\le4=\int_X|f|\,dm\), and the inequality is strict. Check: the positive contribution is \(2\), while the negative contribution is \(-2\); they cancel in \(\int f\,dm\), but both contribute positively to \(\int|f|\,dm\).
Worked Example: A Complex-Valued Function on a Finite Space
Let \(X=\{a,b\}\), let \(\mathcal{M}\) be the set of all subsets of \(X\), and let \(\mu\) be counting measure, so each singleton has measure \(1\). Define \(f(a)=1+i\) and \(f(b)=-i\). Compute the integral and the integral of the magnitude.
For counting measure, integration over this finite set is summation. Hence $$ \int_X f\,d\mu=(1+i)+(-i)=1, \qquad \left|\int_X f\,d\mu\right|=|1|=1. $$ The pointwise magnitudes are \(|f(a)|=|1+i|=\sqrt{1^2+1^2}=\sqrt2\) and \(|f(b)|=|-i|=1\). Consequently, $$ \int_X|f|\,d\mu=\sqrt2+1. $$ Since \(\sqrt2>0\), we have \(1<1+\sqrt2\). The inequality is again strict. Check: summing the two complex values gives \(1\), while summing their magnitudes gives \(1+\sqrt2\), as required by the theorem.
Bounds for Differences of Integrals
Proof. The pointwise triangle inequality for real or complex numbers gives \(|f-g|\le|f|+|g|\). The right-hand side is integrable because \(f\) and \(g\) are integrable and the integral is additive on nonnegative integrable functions. Monotonicity then gives \(\int|f-g|\,d\mu\le\int|f|\,d\mu+\int|g|\,d\mu<\infty\), so \(f-g\) is integrable. By linearity, $$ \int f\,d\mu-\int g\,d\mu=\int(f-g)\,d\mu. $$ Applying the integral triangle inequality to the integrable function \(f-g\) proves the asserted estimate. \(\square\)
The bound has a useful quantitative interpretation: if two integrands are close in \(L^1(\mu)\), their integrals are close. More precisely, for any \(\varepsilon>0\), if \(\|f-g\|_1<\varepsilon\), then $$ \left|\int f\,d\mu-\int g\,d\mu\right|\le\|f-g\|_1<\varepsilon. $$ This conclusion follows directly from the corollary; it does not require an interchange of a limit and an integral.
Worked Example: A Strict Bound for a Difference of Integrals
On \(X=[0,1]\) with Lebesgue measure, define \(f=2\) on \([0,\tfrac12]\) and \(f=-1\) on \((\tfrac12,1]\). Let \(g=0\) everywhere. Compute the difference of the integrals and compare it with \(\int_X|f-g|\,dm\).
Each interval has measure \(\tfrac12\). Therefore $$ \int_X f\,dm=2\cdot\tfrac12+(-1)\cdot\tfrac12=1-\tfrac12=\tfrac12, \qquad \int_X g\,dm=0. $$ The absolute difference of these integrals is \(\left|\tfrac12-0\right|=\tfrac12\). Since \(f-g=f\), its absolute value is \(2\) on the first interval and \(1\) on the second, giving $$ \int_X|f-g|\,dm=2\cdot\tfrac12+1\cdot\tfrac12=1+\tfrac12=\tfrac32. $$ Thus \(\tfrac12\le\tfrac32\), strictly. Check: the integral difference is the signed total \(1-\tfrac12=\tfrac12\), while the bound is the sum of the magnitudes \(1+\tfrac12=\tfrac32\). The only difference between these calculations is the cancellation of the negative contribution in the signed integral.
Worked Example: Equality in the Difference Bound
On \(X=[0,1]\) with Lebesgue measure, define \(f=3\) and \(g=1\) on \([0,\tfrac12]\), and define \(f=-1\) and \(g=-2\) on \((\tfrac12,1]\). Compute both sides of the difference estimate.
The integral of \(f\) is $$ \int_X f\,dm=3\cdot\tfrac12+(-1)\cdot\tfrac12=\tfrac32-\tfrac12=1. $$ The integral of \(g\) is $$ \int_X g\,dm=1\cdot\tfrac12+(-2)\cdot\tfrac12=\tfrac12-1=-\tfrac12. $$ It follows that \(\left|\int_X f\,dm-\int_Xg\,dm\right|=|1-(-\tfrac12)|=\tfrac32\). Pointwise, \(f-g=2\) on the first interval and \(f-g=1\) on the second, so $$ \int_X|f-g|\,dm=2\cdot\tfrac12+1\cdot\tfrac12=1+\tfrac12=\tfrac32. $$ The two sides are equal. Check: \(f-g\) is nonnegative everywhere, so no positive and negative contributions cancel in its integral; indeed \(\int_X(f-g)\,dm=\tfrac32=\int_X|f-g|\,dm\).
A Common Mistake: Replacing the Inequality by Equality
A tempting but incorrect calculation is to move the absolute value through the integral and write \(\left|\int f\,d\mu\right|=\int|f|\,d\mu\) for every integrable \(f\). The first worked example disproves this: its left side is \(0\), whereas its right side is \(4\). The valid statement is an inequality, because signed or complex contributions can cancel before their total is measured. In general, one may use equality only when it has been separately established for the particular function; the triangle inequality alone does not provide it.
Likewise, the difference estimate uses the magnitude of the pointwise difference, \(\int|f-g|\,d\mu\), not the difference of the magnitudes \(\int|f|\,d\mu-\int|g|\,d\mu\). The latter need not be nonnegative and does not measure how far apart the functions are pointwise. The reliable procedure is to combine linearity, \(\int f-\int g=\int(f-g)\), with the theorem applied to \(f-g\).
Check Your Understanding
For each question, identify the measure space and verify that every function whose integral is used is integrable.
- On \([0,2]\) with Lebesgue measure, let \(f=1\) on \([0,1]\) and \(f=-3\) on \((1,2]\). Compute \(\left|\int f\,dm\right|\) and \(\int|f|\,dm\). Is the inequality strict?
- For a complex-valued integrable \(f\) with \(z=\int f\,d\mu\ne0\), explain why choosing \(\alpha=\overline{z}/|z|\) gives \(\operatorname{Re}(\alpha z)=|z|\).
- Suppose \(f,g\in L^1(\mu)\) and \(\int|f-g|\,d\mu=0\). What does the difference bound imply about their integrals?
- On a two-point space with counting measure, choose values for \(f\) at the two points so that \(\left|\int f\,d\mu\right|<\int|f|\,d\mu\), and verify both sides.
- If \(\|f_n-f\|_1\to0\) for integrable functions on a measure space, use the difference estimate to show that \(\int f_n\,d\mu\to\int f\,d\mu\).