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Proposition 2.2.39CB694815500

Proposition 2.2.3.9CB694815500 Let $(X,\mu )$ be a measure space and let $f_ j$, $j\geq 1$, be measurable functions on $X$.

  1. If $f_ j\geq 0$ and $f_ j\uparrow f$ pointwise, then $\int f_ j\, \mathrm d\mu \uparrow \int f\, \mathrm d\mu $.

  2. If $f_ j\geq 0$ for every $j$, then

    \[ \int \varliminf _ j f_ j\, \mathrm d\mu \leq \varliminf _ j\int f_ j\, \mathrm d\mu . \]

  3. If the $f_ j$ are real or complex, converge almost everywhere to a measurable function $f$, and satisfy $|f_ j|\leq g$ for an integrable function $g$, then $f$ is integrable and $\int |f_ j-f|\, \mathrm d\mu \to 0$.

Proof. (1) The integrals $\int f_ j\, \mathrm d\mu $ increase and are bounded by $\int f\, \mathrm d\mu $, so we must show that their limit is at least $\int f\, \mathrm d\mu $. Let $s\leq f$ be a nonnegative simple function and $0<t<1$. The measurable sets $E_ j=\{ f_ j\geq ts\} $ increase with $j$, and their union contains $\{ s>0\} $, since $f(x)\geq s(x)>ts(x)$ there. Applying continuity from below to the sets $E_ j\cap \{ s=a\} $ for the finitely many values $a$ of $s$, we obtain

\[ \lim _ j\int f_ j\, \mathrm d\mu \geq \lim _ jt\int _{E_ j}s\, \mathrm d\mu =t\int s\, \mathrm d\mu . \]

Taking the supremum over $s$ and then letting $t\uparrow 1$ proves (1); infinite integrals are allowed.

(2) The functions $g_ j=\inf _{k\geq j}f_ k$ are measurable, they increase to $\varliminf _ jf_ j$, and $\int g_ j\, \mathrm d\mu \leq \int f_ k\, \mathrm d\mu $ for every $k\geq j$. Applying (1) to $(g_ j)$ gives (2).

(3) Letting $j\to \infty $ in $|f_ j|\leq g$ gives $|f|\leq g$ almost everywhere, so $f$ is integrable. The functions $2g-|f_ j-f|$ are nonnegative almost everywhere, and (2), applied after a modification on a null set, gives

\[ \int 2g\, \mathrm d\mu \leq \varliminf _ j\int \bigl (2g-|f_ j-f|\bigr )\, \mathrm d\mu =\int 2g\, \mathrm d\mu -\varlimsup _ j\int |f_ j-f|\, \mathrm d\mu . \]

Since $\int 2g\, \mathrm d\mu <\infty $, it follows that $\int |f_ j-f|\, \mathrm d\mu \to 0$. $\square$

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