If $f(x)=g(x)$ except for a specific point, how to prove that $\int_{-\infty}^{\infty}f(x) dx =\int_{-\infty}^{\infty}g(x) dx$?

Question

If $f(x)=g(x)$ except for a specific point, how to prove that $$\int_{-\infty}^{\infty}f(x) dx =\int_{-\infty}^{\infty}g(x) dx$$

I know that the integration of a specific point equals to zero, but I look for mathematical proof.

Answer 1

Assume $f(x)=g(x)$ for all $x\in\mathbb{R}$ apart from a single point $a$. Then, if we denote the Lebesgue measure by $m$, we’ve got $$\int\limits_{\mathbb{R}\setminus\{a\}}f(x)\,dm(x)=\int\limits_{\mathbb{R}\setminus\{a\}}g(x)\,dm(x)$$ Now, $\mathbb{R}=(\mathbb{R}\setminus\{a\})\cup\{a\}$, and this union is disjoint. Furthermore, $m(\{a\})=0$, and therefore, since integrating over a $m$-null set yields $0$, we’ve got $$\int\limits_{\{a\}}f(x)\,dm(x)=0=\int\limits_{\{a\}}g(x)\,dm(x)$$ So by additivity of integrals, $$\int\limits_{-\infty}^\infty f(x)\,dx=\int\limits_\mathbb{R}f(x)\,dm(x)=\int\limits_{\mathbb{R}\setminus\{a\}}f(x)\,dm(x)+\int\limits_{\{a\}}f(x)\,dm(x)=\int\limits_{\mathbb{R}\setminus\{a\}}f(x)\,dm(x)=\\ \int\limits_{\mathbb{R}\setminus\{a\}}g(x)\,dm(x)=\int\limits_{\mathbb{R}\setminus\{a\}}g(x)\,dm(x)+\int\limits_{\{a\}}g(x)\,dm(x)=\int\limits_\mathbb{R}g(x)\,dm(x)=\int\limits_{-\infty}^\infty g(x)\,dx$$