$L^1(X)$, delta epsilon measure proof Let $f \in L^1(X)$ with $f \ge 0$. We know that $$\nu(E) := \int_E f\,d\mu$$defines a measure on $\Sigma$. How do I show that for every $ \epsilon > 0$ there exists $\delta > 0$ so that for any $E \in \Sigma$ the property $\mu(E) < \delta$ implies $\nu(E) < \epsilon$?
 A: This is the proof given in Stein's text.
Let $f_n:= f \cdot \chi_{\{f \le n\}}$ (in other words, $f_n$ equals $f$ whenever $f$ is less than $K$, and is zero elsewhere). Then the $(f_n)_{n \ge 1}$ form an increasing sequence, so by the monotone convergence theorem, $$\lim_{n \to \infty} \int f_n \mathop{d\mu} = \int f \mathop{d\mu}.$$
Fix $\epsilon>0$. There exists an integer $N$ such that
$$\int f \mathop{d\mu}-\int f_N \mathop{d\mu}<\epsilon/2.$$
So, for any set $E$ such that $\mu(E) \le \delta := \epsilon/(2N)$,
\begin{align*}
\int_E f\mathop{d\mu}&=\left(\int_E f \mathop{d\mu}-\int_E f_N \mathop{d\mu}\right)+\int_E f_N \mathop{d\mu}\\
&<\epsilon / 2 + \int_E N \mathop{d\mu}\\
&\le \epsilon/2 + N\mu(E) & f_n \le N\\
&< \epsilon.
\end{align*}
A: For this problem we do not actually need $f \ge 0$. Recall that a sequence in a metric space $Y$ converges to a point $p$ in a metric space if and only if every subsequence has a subsequence converging to $p$. Here our metric space is $Y = \mathbb{R}$, our point is $p=0$, and our sequence will be the integrals of $\int \chi_{E_n}$, where $E_n$ sequence of sets with measure tending to $0$. Since any subsequence converges in measure to the zero function $0$, it has a subsequence $f_{\chi_{E_{n_k}}}$ that converges almost everywhere to $0$, dominated by $|f| \in L^1$, so that as $k \to \infty$,$$\int_{E_{n_k}} f\,d\mu = \int_X f_{\chi_{E_{n_k}}}\,d\mu \to 0.$$By the fact we mentioned at the beginning, this implies the result.
Notice that equivalence classes of measurable sets, where the equivalence is by having a null set as the symmetric difference, embed into $L^1(X,\mu)$ by their characteristic functions. Let $[E]$  denote the equivalence class of a measurable set $E$. This result implies that for any $f \in L^1(X,\mu)$, the map $[E] \to \int_E f$ is uniformly continuous for any $f \in L^1$.
