Originally published at 狗和留美者不得入内. You can comment here or there.

In undergrad, I learned these three major theorems for Lebesgue integration but never understood them. I was never able to actually re-prove them independently. However, I succeeded now with all three after reading the gist of them a week or two ago. I don’t them I will ever forget them (by which I mean I will always be able to reconstruct them), though that is of course much easier said than done. I will write them up now to confirm.

**Proposition 1** Assuming countable additivity of , *Proof*: We leave this to the reader.

**Proposition 2** For a real measurable function and any real number , the following are equivalent.

- is measurable.
- is measurable.
- is measurable.
- is measurable.

*Proof*: Left to the reader.

**Proposition 3** For any Lebesgue measurable non-negative function , is countably additive on Lebesgue measurable sets.

*Proof*: We leave this to the reader.

**Theorem 1** **(Monotone convergence theorem)** Let be a sequence of measurable functions such that for all and pointwise on set . Then, is measurable, with

*Proof*: The inequality is obvious from monotonicity. The hard part is developing a technique to prove the other way round. For that, the heuristic is to take a monotonic sequence converging for the right hand side and associate it directly with a sequence converging to the right hand side that is bounded above by with it at every index. For the right hand side, the natural choice is a sequence of simple functions such that

which exists by definition of Lebesgue integral. For any fixed , we can let

The definition of simple function along with Proposition 2 tells us that is measurable for all . since by pointwise convergence, for every , there exists an such that . By Proposition 3, we have for all

Taking the limit of this gives us

Since this holds for all , it must hold for as well, which completes our proof.

**Lemma 1 (Fatou’s lemma)** For a sequence of measurable non-negative functions ,

*Proof*: By definition of , for all , from which follows

is the pointwise limit of a monotonically non-decreasing sequence. Thus, by the monotone convergence theorem along with taking the limit of ,

This completes our proof.

**Theorem** **2** **(Dominated converged theorem)** Suppose sequence of measurable functions pointwise and there exists measurable such that for all . Then is measurable and

*Proof*: By the monotone convergence theorem and Fatou’s lemma, we have

Seeing that proving

suffices via the squeeze theorem, we also notice that

which gives

Multiplying this by yields the lower bound of , which completes our proof. We note that in this proof we have in some sense implicitly assumed decomposition of into and .