Agreement of the Riemann and Lebesgue Integrals for Continuous Functions on a Closed Interval

Let $a<b$ be real numbers, let $[a,b]$ be the closed interval determined by $a$ and $b$, and let $(\mathbb{R},d_{\mathbb{R}})$ be the real line, that is, $\mathbb{R}$ equipped with the absolute value metric. Let $h:[a,b]\to\mathbb{R}$ be continuous on $[a,b]$, as a map from the subset $[a,b]$ of $(\mathbb{R},d_{\mathbb{R}})$ into $(\mathbb{R},d_{\mathbb{R}})$. Define the zero extension $\tilde{h}:\mathbb{R}\to\mathbb{R}$ by $\tilde{h}(x)=h(x)$ for $x\in[a,b]$ and $\tilde{h}(x)=0$ otherwise. Then the following hold.

Claim 1. $h$ is Riemann integrable on $[a,b]$, by Continuous Functions on Compact Intervals are Riemann Integrable.

Claim 2. $\tilde{h}$ is measurable with respect to the Borel $\sigma$-algebra and integrable with respect to Lebesgue measure $\lambda$.

Claim 3. The two integrals agree:

$$\int_{\mathbb{R}}\tilde{h}\,d\lambda=\int_{a}^{b}h(x)\,dx,$$

where the right-hand side is the Riemann integral of claim 1.