Let $f$ be a nonnegative, measurable and integrable function on measure space $(\mathbb{R},X,\mu)$ with Lebesgue measure $\mu$. Then, is the following true:
$$\forall \epsilon>0\exists E(\mu(E)<+\infty):\int_X f d\mu\le\int_E f d\mu+\epsilon$$
I think yes, but am unable to prove it. I think it is related to uniform integrability of $f$ or the simple approximation of integrable functions. Any hints. Thanks beforehand.
On
Alternative proof using the definition of Lebesgue integral, which is the supremum among the integral of all simple functions $\varphi$ satisfying $\varphi\le f$. For all $\epsilon>0$, there exists a simple function $$ \psi=\sum_{i=1}^m \alpha_{S_i}1_{S_i}, $$ (here the $1_{S_i}$ denote the indicative function on $S_i$) such that $\sum_{i=1}^m\mu(S_i)<\infty$ and $\psi\le f$, satisfying $$ \int_X fd\mu\le\int_X\psi d\mu+\epsilon $$ Let $E=\cup_{i=1}^m S$, then $\mu(E)<\infty$ and $$ \int_X\psi d\mu\le\int_Efd\mu. $$
By the monotone convergence theorem, $\int_{[-n,n]}fd\mu=\int_{\mathbb{R}} f\cdot \mathbb{1}_{[-n,n]}d\mu \to \int_{\mathbb{R}} fd\mu$.
Therefore, given $\epsilon>0$, there exists $N$ such that $\int_{[-N,N]}fd\mu>\int_{\mathbb{R}} fd\mu-\epsilon$.