# Lebesgue Integral

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