Question

In: Advanced Math

State and prove Lebesgue Dominated Convergence theorem.

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Expert Solution

Solution :

$\mathrm{\bold{Statement}\::\quad\:Lebesgue\:Dominated\:Convergence\:Theorem\:}$

$\mathrm{Let\:E\:\subseteq \:R^{d}\:be\:given,\:and\:\:suppose\:that\:}f_{n}:\:\mathrm{E\rightarrow [-\infty ,\:\infty ]\:for\:}n\in \mathbb{N}.\mathrm{ \:\:Suppose\: also \:that\: the\: functions\: }f_{n}(x)\:$ $\mathrm{converge\:pointwise\:almost\:everywhere,\:i.e.,\:}$

$\lim _{n\to \infty }\:f\left(x\right)=f\left(x\right)\mathrm{\:for\:a.e.\:}x\:\in \:E.\:$

$\mathrm{If\:there\:is\:a\:single\:function\:}g\mathrm{\:such\:that\::}$

$\mathrm{(a)\quad\:}\left | f_{n}(x) \right |\leq g(x) \:\mathrm{\:a.e.\:for\:every\:}n,\:\mathrm{and}$

$\mathrm{(b)\quad\:}g\mathrm{\:\:is\:integrable,\:i.e,\:\int _{E}}\left | g \right |< \infty$

$\mathrm{then}$

$\lim _{n\to \infty }\:\int _E\:f_{n}=\int _E\lim _{n\to \infty }f_{n}=\int _Ef$

$\mathrm{In\:reality,\:even\:more\:is\:true\:in\:this\:case\::\:}f_{n}\mathrm{ \:converges\: to\: }f\mathrm{\:in\:}L^{-1}\:\mathrm{-norm,\:i.e.,\:}$

$\lim _{n\to \infty }\left \| f-f_{n} \right \|_{1}=\lim _{n\to \infty }\int_{E}\left | f-f_{n}\right |=0.$

$\mathrm{\bold{Proof. }}$

$\mathrm{We\:will\:give\:the\:proof\:for\:nonnegative\:\:}f_{n}\mathrm{\:\:only,\:but\:it\:can\:be\:extended\:to\:general\:\:}f.$

$\mathrm{Suppose\:that\:\:}f_{n}\:\geq \:0\mathrm{\:\:a.e.\:\:Then\:by\:Fatou's\:Lemma,}$

$\int _Ef=\int _E\lim _{n\to \infty }\mathrm{inf\:}f_{n}\leq \lim _{n\to \infty }\mathrm{inf}\int _Ef_{n}.$

$\mathrm{Further,\: since \:\:}g-f_{n}\geq 0\mathrm{\:\:a.e.,}$

$\mathrm{\:we\:can\:apply\:Fatou's\:Lemma\:to\:the\:functions\:\:}g-f_{n},\mathrm{\:\:to\:obtain}$

$\int _Eg-\int _Ef=\int _E(g-f)$

$=\int _E\:\lim _{n\to \infty }\mathrm{inf}(g-f_{n})$

$\leq\:\lim _{n\to \infty } \mathrm{inf}\int _E(g-f_{n})\quad\:\mathrm{(by\:Fatou's\:Lemma)}$

$=\:\lim _{n\to \infty } \mathrm{inf}(\int _Eg-\int _Ef_{n})$

$=\int _Eg-\:\lim _{n\to \infty } \mathrm{sup}\int _Ef_{n}.$

$\mathrm{Therefore,}$

$\int _Ef\leq\lim _{n\to \infty } \mathrm{inf}\int _Ef_{n}\leq\lim _{n\to \infty } \mathrm{sup}\int _Ef_{n}\leq \int _Ef.$

$\mathrm{Hence,\quad}$

$\lim _{n\to \infty }\int _Ef_{n}\:\mathrm{\:exists\:and\:equals\:}\int _Ef.$

Colby Messinger answered 1 year ago

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Question

State and prove Lebesgue Dominated Convergence theorem.

Solutions

Expert Solution

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