Let $f=0$ over the real line, $\mathbb R$, with Lebesgue measure $\mu$. I'm uncertain about how to compute the Lebesgue integral, $\int fd\mu=\int 0d\mu.$ Is it $0$?
In this case, $f=0$ is a simple function. So by definition, $\int fd\mu=0\mu(\mathbb R)=0\cdot \infty.$ Do we simply define this to be zero for Lebesgue integral?
I thought about countable additivity of the measure, i.e. $\mathbb R=\bigcup_{n=0}^\infty E_n$, where $E_n\triangleq (-n-1,-n] \cup [n,n+1),$ and $\mu(E_n)=2.$ So $0\mu(\mathbb R)=0\sum_{n=0}^\infty \mu(E_n)=0\sum_{n=0}^\infty 2.$ But $\sum_{n=0}^\infty 2$ is a divergent series, does the usual distributive law apply here? I.e. can we say $0\sum_{n=0}^\infty 2=\sum_{n=0}^\infty 0\cdot 2=\sum_{n=0}^\infty 0=0?$
Even if so, it still boils down to a definition, isn't it? I.e. Why should $0\sum_{n=0}^\infty 2=\sum_{n=0}^\infty 0\cdot 2?$ Unless we define $0\cdot \infty = 0.$ Or we make the distributive law hold in this case, and $0\cdot \infty = 0$ follows as a consequence.
Did I miss something? Are there other ways to justify what $\int 0d\mu$ should be? Thanks a lot!