Suppose { f n } is a sequence of (Lebesgue) measurable functions on R d with each f n ≥ 0. If f n ( x ) → f ( x ) for a.e. x, then by Fatou's lemma, we have ∫ f ≤ lim inf n → ∞ ∫ f n .Its proof begins with a non-negative function g that is bounded and supported on a set E ⊆ R d of finite measure with g ≤ f. If we set g n = min { g , f n }, then each g n is a measurable function and supported on E. Furthermore, g n ( x ) → g ( x ) for a.e. x. By the bounded convergence theorem, ...The above material is quoted from the book by Stein and Shakarchi with some minor changes, and I wonder why g n → g almost everywhere. I wish I could offer you some useful ideas, but unfortunately I know nothing, which is why I'm here. I would much appreciate it if you could do me a favor. Thank you.

Question

Suppose   {      f    n    } is a sequence of (Lebesgue) measurable functions on             R        d   with each       f    n    ≥  0. If       f    n    (  x  )  →  f  (  x  ) for a.e.   x, then by Fatou&#039;s lemma, we have  ∫  f  ≤      lim inf          n      →      ∞        ∫      f    n    .Its proof begins with a non-negative function   g that is bounded and supported on a set   E  ⊆            R        d   of finite measure with   g  ≤  f. If we set       g    n    =  min  {  g  ,      f    n    }, then each       g    n   is a measurable function and supported on   E. Furthermore,       g    n    (  x  )  →  g  (  x  ) for a.e.   x. By the bounded convergence theorem, ...The above material is quoted from the book by Stein and Shakarchi with some minor changes, and I wonder why       g    n    →  g almost everywhere. I wish I could offer you some useful ideas, but unfortunately I know nothing, which is why I&#039;m here. I would much appreciate it if you could do me a favor. Thank you.

stormsteghj · Accepted Answer

Given   x  ∈            R      d      , we will look at two cases: (You may try writing out these arguments in a formal manner).(i) If   f  (  x  )  =  g  (  x  ), then       f    n    (  x  ) converges to   g  (  x  ). This completes this case as       g    n    (  x  )  =      f    n    (  x  ) or       g    n    (  x  )  =  g  (  x  ) for each   n  ∈      N  .(ii) If   f  (  x  )  &amp;gt;  g  (  x  ), then as       f    n    (  x  ) converges to   f  (  x  ), for large   n,       f    n    (  x  )  &amp;gt;  g  (  x  ), thus       g    n    (  x  )  =  g  (  x  ). So, again,       g    n    (  x  )  →  g  (  x  ).

Laylah Mora · Accepted Answer

The reason is because the function   min  :      R    ×      R    →      R   is continuous. This follows from writing   min  (  a  ,  b  )  =            a      +      b        2    −                    |            a      −      b              |              2  , or from a direct argument by cases (  a  &amp;lt;  b,   a  =  b).Thus in your case, since       f    n    →  f a.e., it follows that   min  (  g  ,      f    n    )  →  min  (  g  ,  f  )  =  g a.e.

Suppose <mo fence="false" stretchy="false">{ f n </msub> <mo fence="false" s

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