This post talks about the justification of convolution from measure theory.

Measure Theory

We first introduce the following three theorems.

Theorem 1:

Suppose f is a measurable function on \mathbb{R}^{d_1}. Then the function \tilde{f}(x,y) = f(x) is measurable on \mathbb{R}^{d_1} \times \mathbb{R}^{d_2}.

Theorem 2:

If f is a measurable function on \mathbb{R}^d, then the function \tilde{f}(x,y) = f(x-y) is measurable on \mathbb{R}^d \times \mathbb{R}^d.

The theorems above ensure the first step of the justification later. Moreover, one classic version of the well-known Fubini’s Theorem will be stated.

Fubini’s Theorem:

We first rewrite

\displaystyle \mathbb{R}^d = \mathbb{R}^{d_1} \times \mathbb{R}^{d_2} \quad \text{where } d= d_1+d_2 \text{ and } d_1,d_2 \geq 1

Suppose f(x,y) is a non-negative measurable function on \mathbb{R}^d \times \mathbb{R}^d. Then for almost every y \in \mathbb{R}^{d_2}

    1. The slice f^y is measurable on \mathbb{R}^{d_1}
    2. The function defined by the following is measurable on \mathbb{R}^{d_2}

      \displaystyle \int_{\mathbb{R}^{d_1}} f^y(x)dx

    3. Moreover, we have

\displaystyle \int_{\mathbb{R}^{d_2}} \left(\int_{\mathbb{R}^{d_1}}f(x,y)dx \right)dy=\int_{\mathbb{R}^d} f(x,y)dxdy

Now we will approach the topic by the following steps. Suppose that f and g are measurable functions on \mathbb{R}^d.

Step 1:

f(x-y)g(y) is measurable on \mathbb{R}^{2d}.

Proof:

By theorem 2, if f is measurable on \mathbb{R}^d, then f(x-y) is measurable on \mathbb{R}^{2d}. Also, by theorem 1, if  g(x) is measurable on \mathbb{R}^d, then \tilde{g}(x,y) = g(y) is measurable on \mathbb{R}^{2d}. Therefore, f(x-y)g(y) is the product of two measurable functions, and it is indeed measurable.

Step 2:

If f and g are integrable on \mathbb{R}^d, then f(x-y)g(y) is integrable on \mathbb{R}^{2d}.

Proof:

From step 1, the function f(x-y)g(y) is measurable, then so is |f(x-y) ||g(y)|. By Fubini’s Theorem,

    \[ \int_{\mathbb{R}^{2d}} |f(x-y)g(y)|dxdy = \int_{\mathbb{R}^d} |g(y)| \left(\int_{\mathbb{R}^d} |f(x-y)|dy\right) dx = ||f||_{L^1}||g||_{L^1} \ \textless \ \infty\]

and the function is integrable.

Step 3:

Define the convolution of f and g to be

\displaystyle (f*g)(x)=\int_{\mathbb{R}^d} f(x-y)g(y)dy

Then f*g is well defined for almost every x. That is, f(x-y)g(y) is integrable on \mathbb{R}^d for almost every x.

Proof:

Since from step 2, f(x-y)g(y) is integrable on \mathbb{R}^{2d}, then by Fubini’s theorem, f(x-y)g(y) is integrable on \mathbb{R}^{d} for any fixed x, i.e., almost every x.

As a matter of fact, various conditions on f and g can guarantee that f*g is well defined for almost everywhere. For instance, one other than above is that if f is bounded and compactly supported, g can be any locally integrable function. To see how this concept can be approached from different aspects, check out the other two volumes of this series (this post and this post).

Most of the content above is based on Real Analysis by Elias M. Stein abd Rami Shakarchi and Real Analysis by Gerald B. Folland. I also learned a lot from the class MAT206 taught by Professor Adam Jacob. Thanks for his kindly teaching!

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