Proof of A Continuous Compactly Supported Function on $\mathbb{R}^n$ is Bounded and Integrable

Throughout, $d(x,0)=\lVert x\rVert$ for $x\in\mathbb{R}^n$, by claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ together with $x-0=x$ in the real vector space $\mathbb{R}^n$.

By claim 2 of Compact Support on $\mathbb{R}^n$ Means Vanishing Outside a Bounded Set there is a real number $R>0$ with $g(x)=0$ for every $x$ with $\lVert x\rVert>R$; fix any such $R$ and let

be the closed ball of centre $0$ and radius $R$. It contains $0$, so it is nonempty; it is compact by A Closed Euclidean Ball is Convex and Compact; and by claim 3 of Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure it is a Borel set with $\lambda_n(\bar B)<\infty$.

Claim 1. The restriction of $g$ to $\bar B$ is continuous on $\bar B$ by Restriction of a Continuous Map, and Continuous Images of Compact Subsets, so by Extreme Value Theorem on a Compact Subset of a Metric Space it attains a largest value $M_1$ and a smallest value $m_1$ on the nonempty compact set $\bar B$. Put $M=\max\{|M_1|,|m_1|\}$, a real number with $M\ge0$. For $x\in\bar B$ we have $m_1\le g(x)\le M_1$, and $-M\le-|m_1|\le m_1$ as well as $M_1\le|M_1|\le M$, so $-M\le g(x)\le M$ and hence $|g(x)|\le M$ by Absolute Value in an Ordered Field. For $x\notin\bar B$ we have $\lVert x\rVert=d(x,0)>R$, so $g(x)=0$ and $|g(x)|=0\le M$. Hence $M$ is a bound for $g$, which is claim 1.

Claim 2. Since $g$ is continuous from $(\mathbb{R}^n,d)$ to $(\mathbb{R},d)$, claim 3(a) of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets shows that $g$ is measurable with respect to $\mathcal{B}(\mathbb{R}^n)$ and the Borel $\sigma$-algebra of the real line. By Integrable Function and the Lebesgue Integral the positive and negative parts $g^{+}$ and $g^{-}$ are measurable and $|g|=g^{+}+g^{-}$, so $|g|$ is measurable by claim 1 of Linearity and Monotonicity of the Lebesgue Integral.

Let $M$ be a bound for $g$, as provided by claim 1, and let $R>0$ be as above. Then

with the indicator function of $\bar B$: for $x\in\bar B$ this is claim 1, and for $x\notin\bar B$ both sides are $0$ because $\lVert x\rVert>R$. The function $\mathbf{1}_{\bar B}$ is measurable and $\int_{\mathbb{R}^n}\mathbf{1}_{\bar B}\,d\lambda_n=\lambda_n(\bar B)$ by The Integral of an Indicator Function is the Measure of the Set, so claim 1 of Linearity and Monotonicity of the Lebesgue Integral, applied first to the constant $M\in[0,\infty)$ and then to the pointwise inequality above, gives

The right-hand side is a product of two real numbers, hence real and in particular smaller than $\infty$. By the criterion recorded in Integrable Function and the Lebesgue Integral, $g$ is integrable with respect to $\lambda_n$.

Claim 3. Assume $g\ge0$ everywhere and $g(x_1)>0$. Since $g^{-}=0$ and $g^{+}=g$ in this case, the integral of Integrable Function and the Lebesgue Integral coincides with the integral of the nonnegative function $g$ in the sense of Lebesgue Integral of a Nonnegative Measurable Function, and it is finite by claim 2.

Put $c=g(x_1)\,2^{-1}$, a real number with $0<c<g(x_1)$. By Semicontinuity Under Negation and Characterization of Continuity the continuous function $g$ is lower semicontinuous on $\mathbb{R}^n$, so by claim 2 of Semicontinuity via Sublevel and Superlevel Sets the set $\{x\in\mathbb{R}^n: c<g(x)\}$ is open in $(\mathbb{R}^n,d)$. It contains $x_1$, so by Open Subset of a Metric Space there is a real number $\rho>0$ such that the open ball $B(x_1,\rho)$ is contained in it.

Consequently $g(x)\ge c\,\mathbf{1}_{B(x_1,\rho)}(x)$ for every $x$: on $B(x_1,\rho)$ we have $g(x)>c$, and elsewhere the right-hand side is $0\le g(x)$. The ball is a Borel set of positive measure by claim 1 of Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure, so claim 1 of Linearity and Monotonicity of the Lebesgue Integral and The Integral of an Indicator Function is the Measure of the Set give

the final inequality because $c$ is a positive real number and $\lambda_n(B(x_1,\rho))$ is a positive element of $[0,\infty]$, so their product is positive under the conventions of Measure, Measure Space, and Probability Measure.