Question about a property of integrable function

Question

For $(\mathbb{R}^d, \mathcal{M}, m)$, where $m$ here is the usual extension of the exterior measure $m_*$ defined by volume of open balls, if $f \in L^1({\mathbb{R}^d})$, then for every $\epsilon >0$, there exists a set of finite measure $B$ (open ball of instance) such that $$\int_{B^c} f(x) dm < \epsilon$$ My question is: Does this property still hold if we consider a more general case where $(X, \mathcal{A}, \mu)$ a $\sigma-$finite measurable space, and $f$ is a $L^1(X)$ function?

Answer 1

For the general space, if $f \geqslant 0$ then since $f$ is integrable, given $\epsilon > 0$ there exists a simple function $\phi$ such that $0 \leqslant \phi \leqslant f$ and

$$\int_X f- \epsilon/2 \leqslant \int_x \phi \\ \implies \int_X (f - \phi) \leqslant \epsilon/2.$$

Since $\phi$ is simple and assumes a finite number of values, there exists $M> 0$ such that $0 \leqslant \phi(x) \leqslant M$. Take $\delta = \epsilon/(2M)$ and for any measurable $E$ with $\mu(E) < \delta $ we have

$$\int_E f = \int_E(f-\phi) + \int_E \phi \leqslant \epsilon/2 + M \mu(E) < \epsilon.$$

More generally, if $f$ is not nonnegative, separately apply this result to the positive and negative parts of $f$ to obtain

$$\left|\int_E f \right| \leqslant \int_E |f| < \epsilon.$$

Furthermore, with $0 \leqslant \phi \leqslant f$, the simple function $\phi$ is integrable over $X$ and we must have $\mu(B) < \infty$ where $B = \{x \in X: \phi(x) > 0 \}$ is measurable and

$$\int_{X \setminus B} f = \int_{X \setminus B} (f - \phi) + \int_{X \setminus B} \phi = \int_{X \setminus B} (f - \phi) \leqslant \int_{X} (f - \phi) \leqslant \epsilon/2 < \epsilon.$$

Again this result can be generalized for functions $f$ that are not nonnegative.