# Are the following properties about Lebesgue integrable functions true?

Let $$(X,\Sigma,\mu)$$ be a complete measure space. $$E\subseteq X$$. The textbook I'm using missed two statements that I consider true. I want to ask whether those are indeed correct and how to prove it.

1. Let $$f,~g:E\to\overline{\Bbb R}$$ (extended reals) be measurable. If $$f=g$$ a.e and $$f$$ is integrable, then $$g$$ is integrable and $$\int_E f=\int_E g$$.
2. If $$E_1,~E_2,~\cdots,~E_n$$ (finitely many) are disjoint measurable sets, $$E=\cup_{i=1}^nE_i$$ and $$f:E\to\overline{\Bbb R}$$ is integrable over each $$E_i$$($$i=1$$ to $$n$$). Then $$f$$ is integrable on $$E$$.

Yes, 1. is true. Integrability means that $$\int_{\mathbb R}|f|<\infty$$, and $$|f|=|g|$$ a.e., thus $$\int_{\mathbb R}|g|=\int_{\mathbb R}|f|<\infty$$ so $$g$$ is integrable. Now splt $$f\cdot 1_E$$ and $$g\cdot 1_E$$ into positive and negative parts and subtract.
Also 2. is true by writing $$|f|$$ as a sum: $$|f|\cdot 1_{E_1}+\cdots +|f|\cdot 1_{E_n}$$ and applying additivity of the integral.
• Thanks. For 2. Why is $|f|\cdot 1_{E_1}$ (which has domain $E$) integrable? I know that $f|_{E_1}$ is integrable (hypothesis) and $f\cdot 1_{E_1}|_{E\setminus E_1}$ is integrable (zero function), but how to deduce that $|f|\cdot 1_{E_1}$ is integrable? – Eric Jan 5 at 7:50
• Because $\int_{\mathbb R} |f|\cdot 1_E = \sum_{i=1}^n \int_{\mathbb R}|f|\cdot 1_{E_i}$, by additivity of the Lebesgue integral. And thus you are adding finitely many quantities each $<\infty$, so the result is also $<\infty$. Hence integrability... – pre-kidney Jan 5 at 7:53
• Oh I answered a different question than what you asked in that last comment. I think you are asking why I can extend the domain from a smaller set to a larger set (multiplying the $1_E$) without changing integrability. This is actually just by definition of the Lebesgue integral, at least how it is normally introduced: first you define the Lebesgue integral on $\mathbb R$, then by definition the Lebesgue integral over a measurable subset $E$ of $\mathbb R$ is defined by multiplying by $1_E$. – pre-kidney Jan 5 at 7:56