Intuitive explanation of the proof of Monotone Convergence Theorem

Goal (fix this first)¶

You are given:

• a measurable set $X \subset \mathbb{R}$,

• a nondecreasing sequence of measurable functions $f_n \ge 0$,

• their pointwise limit $f = \lim_{n \to \infty} f_n$.

You want to prove:

This splits into two inequalities:

1. $\displaystyle \lim \int f_n \le \int f$ (easy)

2. $\displaystyle \int f \le \lim \int f_n$ (hard)

Everything below is about (2).

Step 1 — The easy inequality¶

Because the sequence is increasing:

By monotonicity of the integral:

So the sequence $\left( \int f_n \right)$ is increasing and bounded above. Hence the limit

exists and satisfies:

This part is straightforward.
Now forget it — the difficulty is the reverse inequality.

Step 2 — Strategy for the hard inequality¶

You want to prove:

Key idea:

Do not compare $f$ directly to the $f_n$.
Compare simple pieces of $f$ to the $f_n$.

This works because the Lebesgue integral is defined from below.

Step 3 — Fix a simple function below $f$¶

Take any positive simple function:

Why this choice?

Because by definition:

So it is enough to prove:

Step 4 — Why introduce $\alpha \in (0,1)$¶

Problem: Even if $\varphi(x) \le f(x)$, it is not true that

Solution: Relax the comparison.

Instead of forcing:

you try:

Why this helps:

For every $x$,

and since $f_n(x) \uparrow f(x)$, the inequality

will eventually hold.

This is the engine of the proof.

Step 5 — Define the sets $E_n$¶

Define:

Interpretation:

$E_n$ is the set of points where $f_n$ is already large enough.

Key properties:

1. Measurable — since $\varphi$ and $f_n$ are measurable.

2. Increasing — because $f_n \le f_{n+1}$.

3. Exhausts the space:

   $$E_n \uparrow X.$$

   (16)

Why does $E_n \uparrow X$?

Fix $x \in X$.

• $\varphi(x) \le f(x)$

• so $\alpha \varphi(x) < f(x)$

• hence there exists $N$ such that for all $n \ge N$,

  $$f_n(x) > \alpha \varphi(x),$$

  (17)

• so $x \in E_n$.

Thus every point eventually belongs to some $E_n$.

Step 6 — Restrict $\varphi$ to $E_n$¶

On the set $E_n$, we have:

Outside $E_n$, we do not care.

So we look at:

Why?

• still a simple function,

• increases to $\varphi$,

• dominated by $f_n$ (up to factor $\alpha$).

Step 7 — Integrate $\varphi \mathbf{1}_{E_n}$¶

Compute:

Now:

• $E_n \uparrow X$,

• so $A_j \cap E_n \uparrow A_j$,

• and Lebesgue measure is continuous from below.

Therefore:

hence:

Step 8 — The key inequality¶

On $E_n$:

Integrate:

Let $n \to \infty$:

• left side → $\alpha \int \varphi$,

• right side → $L$.

So:

Since this holds for every $\alpha \in (0,1)$, let $\alpha \to 1$:

Step 9 — Final conclusion¶

You have shown:

Taking the supremum over all such $\varphi$:

Combined with the first inequality:

you conclude:

Mental reconstruction checklist (memorize this)¶

If you remember only this, you can re-prove everything:

1. Fix a simple $\varphi \le f$

2. Fix $\alpha < 1$

3. Define $E_n = \{ \alpha \varphi \le f_n \}$

4. Observe $E_n \uparrow X$

5. Use continuity of measure on $\varphi \mathbf{1}_{E_n}$

6. Compare $\alpha \varphi$ with $f_n$

7. Let $n \to \infty$, then $\alpha \to 1$

8. Take the supremum over $\varphi$

If you can reproduce these steps without notes, you fully own the Monotone Convergence Theorem.