Let $(\Omega,\mathcal{A},\mu)$ be a $\sigma$-finite measure space, $f:\Omega\rightarrow\mathbb{R}$ measurable. Suppose there is a $c\in\mathbb{R}$ such that for all $X\subset\mathbb{R}$ of finite measure $\mid\int_X f\mid\leq c$ holds. Prove that $f\in L^1(\Omega,\mathcal{A},\mu)$.
Attempt) Since the measure space is $\sigma$-finite, $\Omega=\bigcup\limits_{i=1}^{\infty}X_i$ with $\mu(X_i)<\infty$. Set $Y_n=\bigcup\limits_{i=1}^{n}X_i$. Then $\mu(Y_n)<\infty$ and $\mid\int_{Y_n} f\mid\leq c$. Since $\mid\int_{Y_n} f\mid=\mid\int_{\Omega} f\cdot{\chi}_{Y_n}\mid$ for the characteristic function, ${\chi}_{Y_n}$, $\mid\lim\limits_{n\rightarrow\infty}\int_\Omega f\cdot{\chi}_{Y_n}\mid\leq c.$ Also, $\lim\limits_{n\rightarrow\infty}f\cdot{\chi}_{Y_n}=f
$.
I don't know what to do(what theorems to apply) to take the limit inside the integral.
Best Answer
Indeed, let $X= \bigcup_{n=1}^\infty A_n$ where $\mu(A_n) < \infty$ for all $n \geq 1$ and $A_1 \subseteq A_2 \subseteq \dots$. Then we know that
$$\int_{A_n} f^+ d \mu = \left|\int_{A_n} f^+ d \mu\right| = \left|\int_{A_n \cap \{f \ge 0\}} f d \mu\right| \leq c$$
and thus using monotone convergence theorem, since $f^+ I_{A_n} \nearrow f^+$, we get
$$\int_X f^+ d \mu \leq c$$
Similarly, we see that $f^-$ is integrable and thus $f$ is integrable.