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?
 A: 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, $\int f = +\infty$.
A: 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 $\int 1\,d\lambda\ge+\infty$.
