# Lebesgue Integral of 1 over $\mathbb{R}$

We've been studying Lebesgue integrals in my analysis class and I'm unsure how to evaluate, $$\int_{\mathbb{R}} 1 d\lambda(t).$$ I know it's a simple function, is $$\lambda(\mathbb{R})=+\infty$$? If so, would this integral just be infinity?

• yes, you are correct. Commented Nov 5, 2020 at 21:31

Yes. It is $$+\infty$$ because $$\lambda(\mathbb{R}) = +\infty$$. You may check this without regarding $$1$$ (let's call it $$f$$) as a non-negative simple function and evaluating $$1·\infty$$. Simply regard $$f$$ as a non-negative function. Consider the sequence of simple functions $$\varphi_n = \mathbf{1}_{[-n,n]}$$ (here $$\mathbf{1}_{[-n,n]}$$ stands for the indicator function of $$[-n,n]$$), all of which are $$\leq f$$ everywhere. The integral of $$\varphi_n$$ is $$1\cdot 2n$$ and so $$\int f\geq 2n$$ for all $$n\in\mathbb{N}$$, thus, $$\int f = +\infty$$.
By definition the Lebesgue integral of a non-negative function $$f$$ is, as an extended real number, the least upper bound of the set $$\left\{\sum\limits_{x\in\operatorname{range}\phi}x\lambda(\phi^{-1}(x))\,:\,\phi\text{ has finite range }\land0\le\phi\le f\right\}$$, with the convention that $$0\times (+\infty)=0$$ and $$\text{(not minus infinity)}+\infty=+\infty$$. In your case $$\phi(x):=1$$ has finite range and $$\sum\limits_{x\in\operatorname{range}\phi}x\lambda(\phi^{-1}(x))=1\cdot\lambda(\Bbb R)=+\infty$$ and $$\phi$$ is smaller or equal to the integrand. Therefore $$\int 1\,d\lambda\ge+\infty$$.