Lebesgue integral, MCT, DCT, Fatou

In one sentence

All three convergence theorems answer the same question — when does $\int \lim f_n = \lim \int f_n$ — and each one is examined by deleting the hypothesis that stops the mass escaping.

Why the exam asks it

Because the theorems are easy to state and the hypotheses are easy to lose, and because there are exactly two ways the conclusion fails. Once you can draw both, most options in this topic answer themselves.

The idea

For a non-negative measurable $f$, define $\int f$ as the supremum of $\int s$ over simple measurable $0 \le s \le f$. For general $f$, split into $f^+ - f^-$ and require both parts to have finite integral. So $f \in L^1$ means $\int |f| < \infty$; Lebesgue integrability is absolute by construction, and that single fact drives one of the topic's best questions.

The three theorems

Monotone convergence. If $0 \le f_1 \le f_2 \le \cdots$ and $f_n \to f$ pointwise, then $\int f_n \to \int f$. No domination needed. Monotonicity and non-negativity are the hypotheses, and both matter.

Fatou. For $f_n \ge 0$,

$\int \liminf_n f_n \;\le\; \liminf_n \int f_n .$

An inequality, never an equality, and the direction is worth memorising as "the integral of the limit can only be smaller" — mass can be lost in the limit, never gained.

Dominated convergence. If $f_n \to f$ a.e. and $|f_n| \le g$ for a single $g \in L^1$, then $f \in L^1$ and $\int f_n \to \int f$. The dominating function must be independent of $n$ and must itself be integrable.

The two escapes

Sideways. $f_n = \mathbf{1}_{[n, n+1]}$ on $\mathbb{R}$. For each fixed $x$, eventually $f_n(x) = 0$, so $f_n \to 0$ pointwise. But $\int f_n = 1$ for every $n$. Any dominating $g$ would have to be at least $1$ on $[n,n+1]$ for every $n$, hence at least $1$ on $[1,\infty)$, hence not integrable.

Upwards. $f_n = n \cdot \mathbf{1}_{(0, 1/n]}$ on $(0,1)$. Again $f_n \to 0$ pointwise, and again $\int f_n = n \cdot \tfrac1n = 1$. Here the domain has finite measure, so the sideways escape is unavailable; the mass goes up instead. A dominating function would have to beat $n$ on $(0,1/n]$ for every $n$, so it grows like $1/x$, which is not integrable near $0$.

These two examples also make Fatou strict: $\liminf f_n = 0$, so $\int \liminf f_n = 0$, while $\liminf \int f_n = 1$. The inequality is $0 \le 1$ and there is nothing to close.

Note what MCT says about the second example: nothing, because $f_n$ is not monotone. And what DCT says: nothing, because there is no dominating $g$. Each theorem correctly declines.

Riemann against Lebesgue

Lebesgue's criterion: a bounded $f$ on $[a,b]$ is Riemann integrable exactly when its set of discontinuities has measure zero.

• Riemann integrable $\Rightarrow$ Lebesgue integrable, with the same value.
• The converse fails: $\mathbf{1}_{\mathbb{Q}}$ on $[0,1]$ is discontinuous everywhere, so not Riemann integrable, and it is $0$ almost everywhere, so its Lebesgue integral is $0$.
• Thomae's function is continuous off $\mathbb{Q}$, so its discontinuities have measure zero and it is Riemann integrable, with integral $0$.

Improper Riemann is a different thing again. $\dfrac{\sin x}{x}$ on $(0,\infty)$ has a convergent improper Riemann integral,

$\int_0^{\infty}\frac{\sin x}{x}\,dx = \frac{\pi}{2},$

and is not Lebesgue integrable, because $\int_0^\infty \frac{|\sin x|}{x}\,dx = \infty$ — on each interval $[n\pi, (n+1)\pi]$ the integral of $|\sin x|/x$ is at least a constant times $1/(n+1)$, and the harmonic series diverges. The improper integral converges only through cancellation, and the Lebesgue integral refuses to use cancellation because it is defined absolutely. This is the standard example, and it is the one that catches people who believe Lebesgue is uniformly stronger.

Where intuition breaks

"$f_n \to f$ pointwise, so $\int f_n \to \int f$." Both escape examples. This is the error every theorem in the topic exists to prevent.

"MCT needs a dominating function." It does not; monotone increasing and non-negative is enough, and the limit may well have infinite integral. Conflating the hypotheses of MCT and DCT is the most common mistake here.

"Fatou is an equality." $n\mathbf{1}_{(0,1/n]}$ gives $0 \le 1$. And check the direction — options reverse it.

"Fatou needs $f_n \ge 0$? Surely not." It does. Without non-negativity it fails: take $f_n = -\mathbf{1}_{[n,n+1]}$, and $\liminf$ of the integrals is $-1$ while the integral of the $\liminf$ is $0$, so the inequality points the wrong way.

"Lebesgue integrable, so Riemann integrable." $\mathbf{1}_{\mathbb{Q}}$. The implication runs the other way, on a bounded interval.

"Improper Riemann integrable, so Lebesgue integrable." $\sin(x)/x$. A hypothesis dropped, and a subtle one: the Lebesgue integral requires $|f|$ to be integrable, so any convergence that depends on cancellation is lost.

"$f_n \to f$ a.e. and each $f_n \in L^1$, so $f \in L^1$." Not without domination. The spike example has $f_n \in L^1$ with $\int f_n = 1$; the limit is in $L^1$ here, but the integrals do not converge to it, and in other examples the limit fails to be integrable at all.

"$\int f = 0$, so $f = 0$." Only for $f \ge 0$, and only almost everywhere. Without non-negativity, $f(x) = x$ on $[-1,1]$ integrates to $0$.

The exam's angle

1. Every question here is "does $\int$ commute with $\lim$". Decide that first.
2. If it does not, find the escape. Is the domain infinite (sideways) or is the function unbounded (upwards)? One of the two, essentially always.
3. Match the theorem to its hypothesis. Monotone and non-negative: MCT. Dominated: DCT. Non-negative only: Fatou, and only an inequality.
4. Look for a dominating function explicitly. If you cannot name one, suspect the conclusion fails, and look for the escape.
5. For Riemann questions, count the discontinuities. Measure zero means Riemann integrable; the Dirichlet function fails and Thomae passes, and knowing which is which is often the whole question.

The night before

• MCT: $0 \le f_n \uparrow f$ $\Rightarrow$ $\int f_n \to \int f$. No domination needed.
• Fatou: $f_n \ge 0$ $\Rightarrow$ $\int\liminf f_n \le \liminf\int f_n$. Inequality, and it can be strict.
• DCT: $|f_n| \le g \in L^1$ and $f_n \to f$ a.e. $\Rightarrow$ $\int f_n \to \int f$.
• $\mathbf{1}_{[n,n+1]}$: mass escapes sideways. $n\mathbf{1}_{(0,1/n]}$: mass escapes upwards. Both have integral $1$ and pointwise limit $0$.
• Lebesgue's criterion: bounded, discontinuities of measure zero $\iff$ Riemann integrable.
• $\mathbf{1}_{\mathbb{Q}}$: Lebesgue integrable, not Riemann. Thomae: Riemann integrable.
• $\sin(x)/x$ on $(0,\infty)$: improper Riemann $=\pi/2$, not Lebesgue integrable.
• $f \in L^1$ means $\int|f| < \infty$. The Lebesgue integral never converges by cancellation.