Triangle Inequality for Integrals

From Pointwise Bounds to Integral Bounds

The previous tutorial used monotonicity to compare integrals when one nonnegative function is bounded above by another. The same idea turns the pointwise triangle inequality for real numbers into a useful estimate for functions. For integrable functions, it bounds the integral of the absolute value of a sum by the sum of the individual absolute-value integrals. This is a central estimate whenever terms may cancel or when a sum is too complicated to integrate directly.

Throughout, $(X,\mathcal{F},\mu)$ is a measure space, and functions are real-valued and measurable. As in “Integrable Functions,” an integrable function $f$ has $\int_X |f|\,d\mu<\infty$. We use the Monotonicity of the Lebesgue Integral theorem and the linearity of the integral for integrable functions from earlier in this course. We also use finite additivity for nonnegative integrals.

Definition: For an integrable real-valued function $f$, define its $L^1$ size by $\|f\|_1=\int_X |f|\,d\mu$. This quantity is finite and nonnegative. It is a norm when functions equal almost everywhere are identified; on functions considered pointwise, it may be zero for a function that is nonzero only on a null set.

The familiar inequality $\left|\int_X f\,d\mu\right|\leq\int_X|f|\,d\mu$ was proved in “Integrable Functions.” The result developed here is different: it estimates the $L^1$ size of a sum. The distinction matters because cancellation can make $\int_X(f+g)\,d\mu$ small even when $\int_X|f+g|\,d\mu$ is not.

The Triangle Inequality for Integrals

Theorem (Integral Triangle Inequality): If $f$ and $g$ are integrable real-valued measurable functions on $X$, then $f+g$ is integrable and $$ \int_X |f+g|\,d\mu \leq \int_X |f|\,d\mu+\int_X |g|\,d\mu. $$

Proof. At every $x\in X$, the triangle inequality for real numbers gives $$ |f(x)+g(x)|\leq |f(x)|+|g(x)|. $$ The right-hand side is nonnegative and measurable. By finite additivity of the nonnegative integral and the integrability of $f$ and $g$, $$ \int_X\bigl(|f|+|g|\bigr)\,d\mu = \int_X|f|\,d\mu+\int_X|g|\,d\mu <\infty. $$ Monotonicity of the nonnegative integral therefore yields $$ \int_X|f+g|\,d\mu \leq \int_X\bigl(|f|+|g|\bigr)\,d\mu <\infty. $$ In particular, $f+g$ is integrable. Substituting the finite-additivity identity into the inequality proves the theorem.

The proof uses an inequality between nonnegative functions, so monotonicity applies directly. It does not require $f$ and $g$ to have the same sign, nor does it assume that cancellation is absent. Indeed, the right-hand side remains an upper bound precisely because it ignores any cancellation between the two functions.

Worked Example: A Strict Integral Triangle Inequality

On $X=[0,1]$ with Lebesgue measure, take $f(x)=x-\frac12$ and $g(x)=\frac12$. Then $f+g=x$, so $$ \int_0^1|f(x)+g(x)|\,dx = \int_0^1x\,dx = \frac12. $$ For the first term, split the interval at $1/2$: $$ \int_0^1\left|x-\frac12\right|\,dx = \int_0^{1/2}\left(\frac12-x\right)\,dx + \int_{1/2}^1\left(x-\frac12\right)\,dx = \frac18+\frac18 = \frac14. $$ Also, $\int_0^1|g(x)|\,dx=\frac12$. Thus the theorem gives, and the calculations verify, $$ \frac12 = \int_0^1|f+g|\,dx < \int_0^1|f|\,dx+\int_0^1|g|\,dx = \frac14+\frac12 = \frac34. $$ Here the inequality is strict because $f$ is negative on part of the interval while $g$ is positive there.

Exactly When Does Equality Hold?

The triangle inequality can be strict, as the preceding calculation shows. For real numbers $a$ and $b$, equality in $|a+b|\leq |a|+|b|$ holds exactly when $a$ and $b$ have the same sign or at least one is zero; equivalently, when $ab\geq0$. The corresponding condition for functions is required almost everywhere, because values on a null set do not affect the integral.

Theorem (Equality Condition for the Integral Triangle Inequality): Let $f$ and $g$ be integrable real-valued measurable functions. Then $$ \int_X|f+g|\,d\mu = \int_X|f|\,d\mu+\int_X|g|\,d\mu $$ if and only if $f(x)g(x)\geq0$ for almost every $x\in X$.

Proof. Define $$ h(x)=|f(x)|+|g(x)|-|f(x)+g(x)|. $$ The pointwise triangle inequality shows that $h\geq0$. The function is measurable, and $h\leq |f|+|g|$, whose integral is finite. Thus $h$ is integrable. By linearity and finite additivity, $$ \int_Xh\,d\mu = \int_X|f|\,d\mu+\int_X|g|\,d\mu-\int_X|f+g|\,d\mu. $$ Consequently, equality of the two sides in the theorem is equivalent to $\int_Xh\,d\mu=0$. By the Zero Integral Criterion, this holds exactly when $h=0$ almost everywhere.

It remains to interpret the pointwise condition $h(x)=0$. For real numbers $a,b$, if $ab\geq0$, then $a$ and $b$ have the same sign or one is zero, so $|a+b|=|a|+|b|$. If $ab<0$, both numbers are nonzero with opposite signs, and $$ |a+b|=\bigl||a|-|b|\bigr|<|a|+|b|. $$ Thus $|a|+|b|-|a+b|=0$ exactly when $ab\geq0$. Applying this fact at each point shows that $h=0$ almost everywhere exactly when $fg\geq0$ almost everywhere. This proves both directions.

Worked Example: Equality for Functions with Disjoint Supports

On $[0,1]$, let $f=\mathbf{1}_{[0,1/2]}$ and $g=-\mathbf{1}_{(1/2,1]}$. At each point, at least one of $f$ and $g$ is zero, so $fg=0$ everywhere. Their nonzero parts do not overlap, and $$ \int_0^1|f|\,dx=\frac12, \qquad \int_0^1|g|\,dx=\frac12. $$ The sum equals $1$ on $[0,1/2]$ and $-1$ on $(1/2,1]$. Hence $$ \int_0^1|f+g|\,dx = \frac12+\frac12 = 1 = \int_0^1|f|\,dx+\int_0^1|g|\,dx. $$ The functions do not need to be nonnegative everywhere for equality: it is enough that they do not have opposite, nonzero signs at the same point, apart from a null set.

Worked Example: Complete Cancellation

On $[0,1]$, let $f(x)=1$ and $g(x)=-1$. Then $f+g=0$ at every point, while $$ \int_0^1|f+g|\,dx=0, \qquad \int_0^1|f|\,dx+\int_0^1|g|\,dx=1+1=2. $$ Here $fg=-1<0$ everywhere, so the equality condition fails everywhere and the triangle inequality is strict. This example also shows why one cannot replace $\int|f+g|$ by $\int|f|+\int|g|$ as an identity.

Finite Sums and Local Estimates

Repeated application gives the corresponding estimate for any finite number of integrable functions. The result is useful when an expression is built from several terms, since each term can be estimated separately.

Theorem (Triangle Inequality for Finite Sums): If $f_1,\ldots,f_n$ are integrable real-valued measurable functions, then their sum is integrable and $$ \int_X\left|\sum_{k=1}^n f_k\right|\,d\mu \leq \sum_{k=1}^n\int_X|f_k|\,d\mu. $$

Proof. We use induction on $n$. For $n=1$, the assertion is the identity $\int_X|f_1|\,d\mu=\int_X|f_1|\,d\mu$. Suppose it holds for $n$ integrable functions. Their sum $F_n=\sum_{k=1}^n f_k$ is integrable by the induction hypothesis, and $f_{n+1}$ is integrable by assumption. The Integral Triangle Inequality applied to $F_n$ and $f_{n+1}$ gives $$ \int_X|F_n+f_{n+1}|\,d\mu \leq \int_X|F_n|\,d\mu+\int_X|f_{n+1}|\,d\mu \leq \sum_{k=1}^{n+1}\int_X|f_k|\,d\mu. $$ The first inequality also shows that $F_n+f_{n+1}$ is integrable. This proves the assertion for $n+1$, and induction completes the proof.

The estimate also applies on a measurable subset. If $E\in\mathcal{F}$, then $f\mathbf{1}_E$ and $g\mathbf{1}_E$ are integrable, since $|f\mathbf{1}_E|\leq|f|$ and $|g\mathbf{1}_E|\leq|g|$. Applying the Integral Triangle Inequality to these two functions gives $$ \int_E|f+g|\,d\mu \leq \int_E|f|\,d\mu+\int_E|g|\,d\mu. $$ This localized form is handy when only part of the space is relevant.

Worked Example: Estimating a Three-Term Sum

On $[0,1]$, consider $f_1(x)=1$, $f_2(x)=x$, and $f_3(x)=-2x$. The finite-sum inequality gives $$ \int_0^1|f_1+f_2+f_3|\,dx \leq \int_0^1|f_1|\,dx+\int_0^1|f_2|\,dx+\int_0^1|f_3|\,dx. $$ The right-hand side is $$ 1+\frac12+2\int_0^1x\,dx = 1+\frac12+1 = \frac52. $$ In fact, $f_1+f_2+f_3=1-x$, which is nonnegative on $[0,1]$, and $$ \int_0^1|1-x|\,dx = \int_0^1(1-x)\,dx = 1-\frac12 = \frac12. $$ Thus the estimate is valid but not sharp here; the separate bounds discard cancellation between the second and third terms.

How to Use the Inequality Carefully

The triangle inequality is a bound, not an identity. Equality depends on the signs of the summands at almost every point, not merely on whether their integrals have the same sign. For two functions, equality holds exactly when their product is nonnegative almost everywhere. If they have opposite nonzero signs on a set of positive measure, the pointwise inequality is strict there, and the equality condition fails.

Key takeaway: Pointwise triangle inequalities and monotonicity yield $\int_X|f+g|\,d\mu\leq\int_X|f|\,d\mu+\int_X|g|\,d\mu$. Equality holds exactly when the summands have the same sign or one is zero almost everywhere.

Check Your Understanding

Use the integral triangle inequality and its equality condition to answer these questions.

Which pointwise inequality is integrated to prove the Integral Triangle Inequality?
Why does the proof also establish that $f+g$ is integrable?
State the necessary and sufficient almost-everywhere condition for equality in the two-function inequality.
For $f=1$ and $g=-1$ on $[0,1]$, calculate both sides of the inequality and explain why equality fails.
How does the inequality for a measurable subset $E$ follow from the whole-space result?