Fatou's Lemma and Dominated Convergence Theorem

Fatou’s Lemma

Statement

For a sequence of nonnegative measurable functions $\{f_n\}\subset L^+$,

\[\int\liminf_{n\to\infty}f_n \le\liminf_{n\to\infty}\int f_n.\]

Proof

Define the tail infimum by

\[g_n(x)=\inf_{k\ge n}f_k(x)\]

Then $0\le g_n\le g_{n+1}$, each $g_n$ is measurable, and $g_n\uparrow\liminf f_n$. MCT gives

\[\int\liminf_{n\to\infty}f_n =\int\lim_{n\to\infty}g_n =\lim_{n\to\infty}\int g_n.\]

Since $g_n\le f_k$ for every $k\ge n$, monotonicity of the integral gives

\[\int g_n\le\int f_k\quad(k\ge n), \qquad \int g_n\le\inf_{k\ge n}\int f_k.\]

Therefore,

\[\int\liminf_{n\to\infty}f_n =\lim_{n\to\infty}\int g_n \le\lim_{n\to\infty}\inf_{k\ge n}\int f_k =\liminf_{n\to\infty}\int f_n. \qquad\square\]

Proof idea / Assumptions

The tail infima increase even when the original sequence is neither convergent nor monotone. Applying MCT to this sequence is the key step. Nonnegativity and measurability remain essential, and the inequality may be strict.

Dominated Convergence Theorem

Statement

Suppose a sequence of measurable functions $\{f_n\}$ satisfies $f_n\to f$ almost everywhere, and there is some $g\in L^1$ such that $|f_n|\le g$ almost everywhere for every $n$. Then $f\in L^1$ and

\[\lim_{n\to\infty}\int f_n=\int f.\]

Proof

From $|f_n|\to|f|$ almost everywhere and $|f_n|\le g$, we obtain $|f|\le g$ almost everywhere. Therefore,

\[\int|f|\le\int g<\infty, \qquad f\in L^1.\]

Since $-g\le f_n\le g$, apply Fatou’s Lemma to $g+f_n$ and $g-f_n$.

Step 1: Fatou for $g+f_n$

Since $g+f_n\to g+f$ almost everywhere,

\[\int(g+f)\le\liminf_{n\to\infty}\int(g+f_n).\]

Because $g\in L^1$, the finite constant $\int g$ can be separated and subtracted.

\[\int g+\int f \le\int g+\liminf_{n\to\infty}\int f_n, \qquad \int f\le\liminf_{n\to\infty}\int f_n. \tag{1}\]

Step 2: Fatou for $g-f_n$

Since $g-f_n\to g-f$ almost everywhere,

\[\int(g-f)\le\liminf_{n\to\infty}\int(g-f_n),\] \[\int g-\int f \le\liminf_{n\to\infty}\left(\int g-\int f_n\right).\]

Using $\liminf(C-a_n)=C-\limsup a_n$ gives

\[\int g-\int f\le\int g-\limsup_{n\to\infty}\int f_n, \qquad \limsup_{n\to\infty}\int f_n\le\int f. \tag{2}\]

From (1), (2), and $\liminf a_n\le\limsup a_n$,

\[\limsup_{n\to\infty}\int f_n \le\int f \le\liminf_{n\to\infty}\int f_n \le\limsup_{n\to\infty}\int f_n.\]

Thus $\lim_{n\to\infty}\int f_n=\int f$. $\square$

Why Domination Is Necessary

On $(0,1)$, let

\[f_n(x)=n\mathbf1_{(0,1/n)}(x)\]

Then $f_n(x)\to0$ for every $x>0$, but

\[\int_0^1 f_n(x)\,dx=1\quad\text{for every }n.\]

Consequently, $\lim_n\int f_n=1\ne0=\int\lim_n f_n$. Integrable domination prevents this kind of mass concentration.

Comparison / What I should remember

Result	Assumptions	Conclusion
MCT	$0\le f_n\uparrow f$, measurable	$\int f_n\to\int f$
Fatou	$f_n\ge0$, measurable	$\int\liminf f_n\le\liminf\int f_n$
DCT	measurable $f_n,f$, $f_n\to f$ a.e., $\lvert f_n\rvert\le g\in L^1$ a.e.	$f\in L^1$, $\int f_n\to\int f$

\[\text{MCT}\longrightarrow\text{Fatou's Lemma}\longrightarrow\text{DCT}\]

The proof ideas to reconstruct are tail infima and $g\pm f_n$. DCT replaces monotonicity with integrable domination.