Proof of The Convolution Integrand is Continuous, Compactly Supported and Integrable

Throughout, $d(z,w)=\lVert z-w\rVert$ by claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, and topological notions refer to the topology of the sets open in $(\mathbb{R}^n,d)$, a topology by Metric Open Sets Form a Topology. On $\mathbb{R}=\mathbb{R}^{1}$ claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ identifies the Euclidean norm of a real number with its absolute value, so by claim 2 there $d(s,t)=|s-t|$, which is the metric on the real line supplied by The Absolute Value Metric on the Real Line and used in Continuity of Sums and Products of Real-Valued Functions on a Metric Space. Put

$$U_1=\{y\in\mathbb{R}^n:x-y\in\Omega\},\qquad U_2=\{y\in\mathbb{R}^n:\lVert y\rVert>\delta\}.$$

Step 1: $U_1$ and $U_2$ are open and cover $\mathbb{R}^n$. Let $y_0\in U_1$. Since $x-y_0\in\Omega$ and $\Omega$ is open there is a real $r>0$ with the open ball $B(x-y_0,r)$ contained in $\Omega$. If $d(y,y_0)<r$ then, since $(x-y)-(x-y_0)=y_0-y$, we get $d(x-y,x-y_0)=\lVert y_0-y\rVert=d(y,y_0)<r$, so $x-y\in\Omega$ and $y\in U_1$. Hence $U_1$ is open.

Let $y_0\in U_2$ and put $r=\lVert y_0\rVert-\delta>0$. If $d(y,y_0)<r$ then claim 6 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, applied to the points $y$ and $y_0-y$ whose sum is $y_0$, gives $\lVert y_0\rVert\le\lVert y\rVert+\lVert y_0-y\rVert<\lVert y\rVert+r$, so $\lVert y\rVert>\lVert y_0\rVert-r=\delta$ and $y\in U_2$. Hence $U_2$ is open.

If $y\notin U_2$ then $\lVert y\rVert\le\delta$, and since $(x-y)-x=-y$ has norm $\lVert y\rVert$ by claim 5 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, the point $x-y$ lies in $\bar B(x,\delta)\subseteq\Omega$; so $y\in U_1$. Thus $U_1\cup U_2=\mathbb{R}^n$.

Step 2: $h_x$ is continuous. On $U_1$ we have $h_x(y)=f(x-y)\rho(y)$ by definition. The map $y\mapsto f(x-y)$ is continuous on $U_1$: given $y_0\in U_1$ and a real $\varepsilon>0$, continuity of $f$ at $x-y_0$ relative to $\Omega$ supplies a real $\theta>0$ such that every $z\in\Omega$ with $d(z,x-y_0)<\theta$ satisfies $|f(z)-f(x-y_0)|<\varepsilon$; and for $y\in U_1$ with $d(y,y_0)<\theta$ the point $z=x-y$ lies in $\Omega$ and satisfies $d(x-y,x-y_0)=d(y,y_0)<\theta$, as computed in step 1. The map $\rho$, being continuous on $\mathbb{R}^n$, is continuous on $U_1$ relative to $U_1$, since a $\theta$ that works for points of $\mathbb{R}^n$ works in particular for points of $U_1$. Hence $h_x$ is continuous on $U_1$ by Continuity of Sums and Products of Real-Valued Functions on a Metric Space.

On $U_2$ we have $h_x(y)=0$: either $x-y\notin\Omega$, and then $h_x(y)=0$ by definition, or $x-y\in\Omega$, and then $h_x(y)=f(x-y)\rho(y)=0$ because $\lVert y\rVert>\delta$ forces $\rho(y)=0$. A constant function is continuous on $U_2$.

Now let $y_0\in\mathbb{R}^n$ and let $\varepsilon>0$ be real. By step 1 the point $y_0$ lies in $U_i$ for some $i\in\{1,2\}$, and $U_i$ is open, so there is a real $r>0$ with $B(y_0,r)\subseteq U_i$. Continuity of $h_x$ on $U_i$ at $y_0$ relative to $U_i$ supplies a real $\theta>0$ such that every $y\in U_i$ with $d(y,y_0)<\theta$ satisfies $|h_x(y)-h_x(y_0)|<\varepsilon$. Let $\theta'$ be the smaller of $r$ and $\theta$; every $y\in\mathbb{R}^n$ with $d(y,y_0)<\theta'$ lies in $U_i$ and satisfies $|h_x(y)-h_x(y_0)|<\varepsilon$. Hence $h_x$ is continuous at $y_0$ relative to $\mathbb{R}^n$, and as $y_0$ was arbitrary, $h_x$ is continuous on $\mathbb{R}^n$.

Step 3: support and integrability. By step 2, $h_x(y)=0$ whenever $\lVert y\rVert>\delta$, so $h_x$ is compactly supported by claim 2 of Compact Support on $\mathbb{R}^n$ Means Vanishing Outside a Bounded Set, applied with $R=\delta$. Claims 1 and 2 of A Continuous Compactly Supported Function on $\mathbb{R}^n$ is Bounded and Integrable now give that $h_x$ is bounded, measurable and integrable with respect to $\lambda_n$. This proves claim 1. The same two claims, applied to $\rho$, which is continuous and vanishes off the same ball and is therefore compactly supported, show that $\rho$ is integrable; hence $\int_{\mathbb{R}^n}|\rho|\,d\lambda_n$ is a real number, by Integrable Function and the Lebesgue Integral.

Step 4: the bound. Let $M\ge0$ be real with $|f(z)|\le M$ for every $z\in\bar B(x,\delta)$, and let $y\in\mathbb{R}^n$. If $\lVert y\rVert>\delta$ then $h_x(y)=0$ and $\rho(y)=0$, so $|h_x(y)|=0=M|\rho(y)|$. If $\lVert y\rVert\le\delta$ then, as in step 1, $x-y\in\bar B(x,\delta)\subseteq\Omega$, so $h_x(y)=f(x-y)\rho(y)$ and $|h_x(y)|=|f(x-y)|\,|\rho(y)|\le M|\rho(y)|$. Thus $|h_x|\le M|\rho|$ pointwise.

Both $|h_x|$ and $M|\rho|$ are integrable with respect to $\lambda_n$: the first because $h_x$ is integrable, the second by claim 2 of Linearity and Monotonicity of the Lebesgue Integral applied to the integrable function $|\rho|$. By claim 2 of that theorem,

$$\Bigl|\int_{\mathbb{R}^n}h_x\,d\lambda_n\Bigr|\le\int_{\mathbb{R}^n}|h_x|\,d\lambda_n\le\int_{\mathbb{R}^n}M|\rho|\,d\lambda_n=M\int_{\mathbb{R}^n}|\rho|\,d\lambda_n,$$

the middle inequality by the monotonicity clause there and the last equality by its linearity clause. This proves claim 2.