Let f be a non-negative integrable function over a set E.Prove that for any $\displaystyle \epsilon > 0$,there exists a set X $\displaystyle \subset$ E such that $\displaystyle m(X)< \infty$ and $\displaystyle \int_E f dm \le \int_X f dm + \epsilon$
Let f be a non-negative integrable function over a set E.Prove that for any $\displaystyle \epsilon > 0$,there exists a set X $\displaystyle \subset$ E such that $\displaystyle m(X)< \infty$ and $\displaystyle \int_E f dm \le \int_X f dm + \epsilon$
Let $\displaystyle f_n:=f1_{X_n}$ where $\displaystyle X_n= E \cap B_m(0,n)$ where $\displaystyle B_m(0,n)=\{ x \in \mathbb{R} ^m : \Vert x \Vert \leq n \}$ that is $\displaystyle f_n(x)=f(x)$ if $\displaystyle x \in X_n$ and $\displaystyle f_n(x)=0$ otherwise. For all $\displaystyle n \in \mathbb{N}$ $\displaystyle \mu (X_n) < \infty$, $\displaystyle 0 \leq f_n \leq f$ for all $\displaystyle n$ and $\displaystyle f_n(x) \rightarrow f(x)$ a.e. in $\displaystyle E$ then by the dominated convergence theorem we have $\displaystyle \int_{E} \ f = \lim_{n \rightarrow \infty} \int_{E} \ f_n$ then for any $\displaystyle \epsilon >0$ there exists an $\displaystyle N$ such that if $\displaystyle n>N$ then (because $\displaystyle f_n \leq f_{n+1}$ for all $\displaystyle n$ and the monotony of the integral) $\displaystyle \int_{E} \ f - \int_{E} \ f_n < \epsilon$.