The Integral of a Convolution

Why the Integral Factors

Convolution combines two functions by integrating their translated product. Its integral over the whole line has a particularly simple form: for integrable functions, it is the product of their separate integrals. The key is to view the convolution as an iterated integral in two variables. For signed or complex-valued functions, Fubini is available only after absolute integrability is checked. For nonnegative functions, Tonelli permits the iterated integrals even when their value is infinite.

Throughout, $m$ denotes Lebesgue measure on $\mathbb{R}$, and $(\mathbb{R},\mathcal{L},m)$ is the complete, sigma-finite Lebesgue measure space. The space $L^1(m)$ consists of equivalence classes, modulo equality $m$-a.e., of real-valued or complex-valued measurable functions $f$ for which $\int_{\mathbb{R}}|f|\,dm<\infty$. We use the convolution convention

(f*g)(x)=\int_{\mathbb{R}} f(x-y)g(y)\,dy.

For $f,g\in L^1(m)$, this integral is finite for $m$-a.e. $x$, and defines an $L^1(m)$ function up to equality a.e. The proof below also verifies the relevant existence and integrability facts. A technical point matters: arbitrary Lebesgue-measurable representatives do not automatically make $(x,y)\mapsto f(x-y)g(y)$ measurable for the raw product sigma-algebra $\mathcal{L}\otimes\mathcal{L}$. We avoid that issue by choosing Borel representatives.

Representative fact: Every real-valued or complex-valued Lebesgue-measurable function on $\mathbb{R}$ that is finite a.e. agrees a.e. with a Borel-measurable function. Indeed, the Lebesgue sigma-algebra is the completion of $\mathcal{B}_{\mathbb{R}}$: each Lebesgue-measurable set differs from a Borel set by a subset of a Borel null set. Apply this to the countably many rational level sets of the real and imaginary parts; outside the union of the resulting null sets, the rational level sets determine the original values. The resulting Borel versions can be set to zero on any Borel null set where they are not finite.

Replacing an $L^1$ representative by a Borel version does not change its integral. It does not change the convolution integral for any fixed $x$ either: changing $f$ on a null set $N$ changes $f(x-y)$ only when $y\in x-N$, a null set, and changing $g$ on a null set changes the integrand only on that null set of $y$-values. Thus the Borel versions can be used to prove identities for the original $L^1$ classes.

The Signed-Integrable Identity

Theorem: Let $f,g\in L^1(m)$, real-valued or complex-valued. Then $f*g\in L^1(m)$, and $$ \int_{\mathbb{R}}(f*g)(x)\,dx = \left(\int_{\mathbb{R}}f(x)\,dx\right) \left(\int_{\mathbb{R}}g(x)\,dx\right). $$ The convolution is defined by the displayed integral for a.e. $x$; the equality is an equality of finite real or complex numbers.

Proof. Choose finite Borel representatives of $f$ and $g$. The map $(x,y)\mapsto x-y$ is continuous, so $(x,y)\mapsto f(x-y)g(y)$ is Borel-measurable on $\mathbb{R}^2$, and hence measurable for the product Borel measure $m\times m$. This is the product measure on $\mathcal{B}_{\mathbb{R}}\otimes\mathcal{B}_{\mathbb{R}}$; we do not assert measurability for arbitrary representatives in $\mathcal{L}\otimes\mathcal{L}$.

The function $(x,y)\mapsto |f(x-y)||g(y)|$ is nonnegative and Borel-measurable. Tonelli's theorem applies on the sigma-finite product space, and translation invariance of Lebesgue measure gives

\int_{\mathbb{R}}\int_{\mathbb{R}}|f(x-y)||g(y)|\,dx\,dy = \int_{\mathbb{R}}|g(y)| \left(\int_{\mathbb{R}}|f(x-y)|\,dx\right)dy = \|f\|_1\|g\|_1 < \infty.

Here $\int_{\mathbb{R}}|f(x-y)|\,dx=\|f\|_1$ for every $y$, by the translation $u=x-y$. The finite absolute double integral permits Fubini's theorem for the real and imaginary parts of the integrand. In particular, for a.e. $x$, the defining integral for $(f*g)(x)$ exists absolutely, and

\int_{\mathbb{R}}|(f*g)(x)|\,dx \le \int_{\mathbb{R}}\int_{\mathbb{R}}|f(x-y)||g(y)|\,dy\,dx = \|f\|_1\|g\|_1.

So $f*g\in L^1(m)$. A further application of Fubini, justified by the same finite absolute integral, gives

\begin{aligned} \int_{\mathbb{R}}(f*g)(x)\,dx &=\int_{\mathbb{R}}\int_{\mathbb{R}} f(x-y)g(y)\,dy\,dx\\ &=\int_{\mathbb{R}}g(y)\left(\int_{\mathbb{R}}f(x-y)\,dx\right)dy\\ &=\int_{\mathbb{R}}g(y)\left(\int_{\mathbb{R}}f(u)\,du\right)dy\\ &=\left(\int_{\mathbb{R}}f(u)\,du\right) \left(\int_{\mathbb{R}}g(y)\,dy\right). \end{aligned}

The translations and the integrals of the chosen representatives agree with those of the original $L^1$ classes, as noted above. This proves the identity. $\square$

Nonnegative Functions and Tonelli

Proposition: Let $f,g:\mathbb{R}\to[0,\infty]$ be Lebesgue-measurable. Choose Borel versions, and use the convention $0\cdot\infty=0$ in their product. Define $(f*g)(x)=\int_{\mathbb{R}}f(x-y)g(y)\,dy$, allowing $+\infty$. Tonelli's theorem gives $$ \int_{\mathbb{R}}(f*g)(x)\,dx = \int_{\mathbb{R}}\int_{\mathbb{R}}f(x-y)g(y)\,dy\,dx = \left(\int_{\mathbb{R}}f\,dm\right)\left(\int_{\mathbb{R}}g\,dm\right) $$ whenever both integrals on the right are finite or both are strictly positive (with $+\infty$ allowed). If either integral is zero, then $f*g=0$ a.e. and its integral is zero; the product $0\cdot\infty$ is otherwise left undefined.

Proof. With Borel versions, $F(x,y)=f(x-y)g(y)$ is a nonnegative Borel-measurable function on $\mathbb{R}^2$. Tonelli applies, without requiring $\int F$ to be finite. For each fixed $y$, translation invariance gives $\int_{\mathbb{R}}f(x-y)\,dx=A$, where $A=\int f\,dm$. If $A<\infty$, the inner integral in $x$ of $F(x,y)$ is $A g(y)$. If $A=\infty$ and $g(y)>0$, that inner integral is $+\infty$; if $g(y)=0$, it is zero under the stated convention. Integrating in $y$ gives $AB$ when $A,B$ are both finite or both positive, where $B=\int g\,dm$. If $A=0$, then $f=0$ a.e.; if $B=0$, then $g=0$ a.e. In either case, Tonelli shows that the double integral, and hence the integral of the convolution, is zero. The Borel-version argument ensures product measurability; null-set changes do not alter the resulting a.e. convolution or its integral. $\square$

The proposition includes the signed theorem's nonnegative $L^1$ case, but its point is broader: nonnegativity lets Tonelli handle an infinite integral. For signed or complex-valued functions, the corresponding argument requires absolute integrability, as checked in the theorem.

Worked Example: Convolution of Two Interval Indicators

Let $f=\chi_{[0,2]}$ and $g=\chi_{[0,1]}$ on $(\mathbb{R},\mathcal{L},m)$. Both are nonnegative Borel functions in $L^1(m)$, with integrals $2$ and $1$. For fixed $x$, the convolution counts the $y\in[0,1]$ for which $x-y\in[0,2]$, or equivalently $y\in[x-2,x]$. The overlap length is

(f*g)(x)= \begin{cases} x,&0\le x\le1,\\ 1,&1\le x\le2,\\ 3-x,&2\le x\le3,\\ 0,&x<0\text{ or }x>3. \end{cases}

Endpoint values agree between adjacent formulas, so there is no ambiguity. Integrating the pieces gives

\int_{\mathbb{R}}(f*g)(x)\,dx = \int_0^1x\,dx+\int_1^2 1\,dx+\int_2^3(3-x)\,dx = \frac12+1+\frac12=2.

Check: the product of the separate integrals is $(\int f\,dm)(\int g\,dm)=2\cdot1=2$, in agreement with the direct computation.

Worked Example: A Signed Convolution with Integral Zero

Let $f=\chi_{[0,1]}-\chi_{[1,2]}$ and $g=\chi_{[0,1]}$. These are real-valued Borel functions in $L^1(m)$. Their integrals are $\int f\,dm=1-1=0$ and $\int g\,dm=1$. Set $T=\chi_{[0,1]}*\chi_{[0,1]}$. The interval-overlap calculation gives $T(x)=x$ on $[0,1]$, $T(x)=2-x$ on $[1,2]$, and $T(x)=0$ off $[0,2]$. Translating the first factor by $1$ shifts this convolution, so

$$ (f*g)(x)=T(x)-T(x-1). $$

The two triangles each have integral $\tfrac12+\tfrac12=1$, so $\int(f*g)\,dm=1-1=0$. This agrees with the theorem's product $0\cdot1=0$. For a pointwise check, at $x=\tfrac12$, $T(\tfrac12)=\tfrac12$ and $T(-\tfrac12)=0$, hence $(f*g)(\tfrac12)=\tfrac12$. The convolution is not the zero function even though its integral is zero.

Worked Example: One-Sided Exponential Densities

On $(\mathbb{R},\mathcal{L},m)$, take $f(x)=e^{-x}\chi_{[0,\infty)}(x)$ and $g(x)=2e^{-2x}\chi_{[0,\infty)}(x)$. They are nonnegative Borel functions in $L^1(m)$. Direct integration gives $\int f\,dm=1$ and $\int g\,dm=2\cdot\frac12=1$. If $x<0$, the convolution is zero. If $x\ge0$, the integrand is nonzero precisely for $0\le y\le x$, and

\begin{aligned} (f*g)(x) &=\int_0^x e^{-(x-y)}2e^{-2y}\,dy\\ &=2e^{-x}\int_0^x e^{-y}\,dy\\ &=2e^{-x}(1-e^{-x}). \end{aligned}

Thus

\int_{\mathbb{R}}(f*g)(x)\,dx = 2\int_0^\infty(e^{-x}-e^{-2x})\,dx = 2\left(1-\frac12\right)=1.

Check: the product of the two separate integrals is $1\cdot1=1$, matching the direct integral.

A Common Pitfall: Confusing Integral Zero with Function Zero

The signed example shows why the convolution identity is an identity of integrals, not a pointwise formula. A tempting but incorrect conclusion is: since $\int f\,dm=0$, the convolution $f*g$ must vanish. In the example above this would give $(f*g)(\tfrac12)=0$, but the direct overlap computation gives $(f*g)(\tfrac12)=\tfrac12$. The correction is that positive and negative parts can cancel after integration: the two triangle terms each have integral $1$, so their difference has integral zero while remaining nonzero at many points.

A second important check is measurability. For arbitrary Lebesgue-measurable representatives, it is not enough to note that $(x,y)\mapsto x-y$ is continuous and immediately claim that $(x,y)\mapsto f(x-y)g(y)$ belongs to $\mathcal{L}\otimes\mathcal{L}$. The composition is Borel-measurable when $f,g$ are Borel, but that does not establish the raw-product assertion for arbitrary Lebesgue representatives. Choose Borel versions first, as in the proof; then Tonelli or Fubini is applied on the Borel product space, with its hypotheses checked.

Case	Tool	Essential condition
Real or complex $f,g\in L^1(m)$	Fubini's theorem	The absolute double integral is finite: $\\|f\\|_1\\|g\\|_1$.
Nonnegative measurable $f,g$	Tonelli's theorem	Nonnegativity and product measurability; the integral may be infinite.
Arbitrary Lebesgue representatives	Choose Borel versions first	Then $f(x-y)g(y)$ is Borel-measurable on $\mathbb{R}^2$.

Check Your Understanding

For each question, state the hypotheses that justify the integral or theorem you use.

For $f=\chi_{[0,1]}$ and $g=\chi_{[0,2]}$, compute $\int(f*g)\,dm$ from the theorem and from the separate integrals.
Explain why the signed $L^1$ proof checks $\int\int |f(x-y)g(y)|\,dx\,dy$ before applying Fubini.
If $f,g\ge0$ are measurable and $\int f\,dm=0$, what can be concluded about $\int(f*g)\,dm$?
For $f=\chi_{[0,1]}-\chi_{[1,2]}$ and $g=\chi_{[0,1]}$, why does $\int(f*g)\,dm=0$ not imply $f*g=0$ a.e.?
Why do Borel representatives resolve the product-measurability issue in the two-variable proof?

Case	Tool	Essential condition
Real or complex \(f,g\in L^1(m)\)	Fubini's theorem	The absolute double integral is finite: \(\\|f\\|_1\\|g\\|_1\).
Nonnegative measurable \(f,g\)	Tonelli's theorem	Nonnegativity and product measurability; the integral may be infinite.
Arbitrary Lebesgue representatives	Choose Borel versions first	Then \(f(x-y)g(y)\) is Borel-measurable on \(\mathbb{R}^2\).

What You'll Learn