Let $a_n$ be a sequence of nonnegative real numbers. Let $E = [1,\infty)$, and $f =a_n$ if $n\leq x<n+1$.Show that $\int_Ef = \sum a_n$

Question

I have the following problem;

Let $\{a_n\}$ be a sequence of nonnegative real numbers. Define the function $f$ on $E = [1,\infty)$ by setting $f(x) = a_n$ if $n\leq x<n+1$. Show that $\int_Ef = \sum_{n=1}^\infty a_n$ .

Can anybody help?

Answer 1

Let $\chi_S$ be the characteristic function of a set $S$. $$ f = 0\chi_{]-\infty,0[} + \sum_{n=1}^\infty a_n \chi_{[n,n+1[} $$ $f$ may not be integrable if $\sum_{n=1}^\infty a_n$ does not converge. If $f$ is integrable, $$\int_{[0,\infty[} f\,d\lambda = \lim_{c \to \infty} \int_{[0,c]} f\,d\lambda $$ By the Monotone Convergence Theorem. Thus its integral is $$ \int_{[0,\infty[} f\,d\lambda = \int \sum_{n=1}^\infty a_n \chi_{[n,n+1[ \cap [0,\infty[}\,d\lambda = \sum_{n=1}^\infty a_n \lambda([n,n+1[) = \sum_{n=1}^\infty a_n $$ as desired.