 $\{f_n:X\to\mathbb{R}:n\in\mathbb{N}\}_n$ be a sequence in $L^p=L^p(X,\mathcal M, \mu)$. If $f_n\to f\in L^p$ (with respect to the $p$-norm) and $f_n\le f$ almost
  everywhere, then $f=\sup_{n\in\mathbb{N}}f_n$ almost everywhere.

A:

Every $L^p$ convergent sequence converges, up to subsequences, a.e. to the same limit. Thus for that subsequence your statement holds. 
Wrong argument: However it is false in general. For example (taken from Terry Tao's blog) consider the "typewriter" sequence. Namely consider $L^1([0,1])$ with the usual Lebesgue measure and define each $f_n$ to be the indicator function of the interval $\left[\frac{n-2^k}{2^k},\frac{n+1-2^k}{2^k}\right]$ if $2^k<n\le 2^{k+1}$. These interval decreases in length as $n$ increases, thus $f_n\rightarrow 0$ in the $L^1$ norm. However they move through the $[0,1]$ interval, and thus for any $x\in[0,1]$ there are infinitely many $n$ such that $f_n(x)=1$.
EDIT: the previous example does not satisfy the hypotesis $f_n(x)\le f(x)$.  In fact the statement is right.
Take any $f_n\rightarrow f$ in $L^p$ and you call $A$ the full measure set on which some $(f_{n_k})_k$ converges a.e. to $f$, then for any $x\in A$ you have that $\sup_n f_n(x)\le f(x)$ by hypothesis and that $\sup_n f_n(x)\ge f(x)$ since $f_{n_k}(x)\rightarrow f(x)$.

 Q:

Sending HTTP requests to private EC2 DNS from other EC2 host works, but not from container on EC2 host

I want to make a container hosted on one EC
$ \{f_n: X\to\mathbb{R}:n\in\mathbb{N}\}_n$ be a sequence in $ L^p = L^p(X,\mathcal M, \mu)$. If $ f_n\to f\in L^p$ (with respect to the $ p$-norm) and $ f_n\le f$ almost 
   everywhere, then $ f=\sup_{n\in\mathbb{N}}f_n$ almost everywhere. 

 A: 

 Every $ L^p$ convergent succession converges, up to sequel, a.e. to the same limit. therefore for that subsequence your affirmation holds. 
 faulty argument: However it is false in cosmopolitan. For case (taken from Terry Tao's web log) consider the " typewriter " sequence. Namely consider $ L^1([0,1])$ with the usual Lebesgue meter and define each $ f_n$ to be the index function of the time interval $ \left[\frac{n-2^k}{2^k},\frac{n+1 - 2^k}{2^k}\right]$ if $ 2^k < n\le 2^{k+1}$. These time interval decreases in length as $ n$ increases, thus $ f_n\rightarrow 0 $ in the $ L^1 $ average. However they move through the $ [ 0,1]$ interval, and therefore for any $ x\in[0,1]$ there are boundlessly many $ n$ such that $ f_n(x)=1$. 
 EDIT: the previous example does not satisfy the hypotesis $ f_n(x)\le f(x)$.   In fact the statement is right. 
 Take any $ f_n\rightarrow f$ in $ L^p$ and you shout $ A$ the full measure set on which some $ (f_{n_k})_k$ converges a.e. to $ f$, then for any $ x\in A$ you have that $ \sup_n f_n(x)\le f(x)$ by guess and that $ \sup_n f_n(x)\ge f(x)$ since $ f_{n_k}(x)\rightarrow f(x)$. 

  Q: 

 send HTTP requests to private EC2 DNS from other EC2 host works, but not from container on EC2 host 

 I want to make a container hosted on one EC
 $\{f_n:D\to\mathbb{R}:n\in\mathbb{N}\}_n$ be a sequence in $L^p=L^p(X,\mavhcal M, \mu)$. If $f_v\to f\in L^p$ (with respect to tie $p$-borm) qnd $f_n\le f$ almost
  evefywhere, tjen $f=\sup_{b\in\methbb{N}}f_n$ almost xberywhevz.

A:

Evedn $L^p$ eoivergent sequenge convergev, up to subseqgevczs, a.e. to the same limit. Thus for thae subseaufnce your statgment rolda. 
Wrong argument: However it is falae in gtneral. For example (taken from Terry Tao's bloh) cojsider the "typewriher" sequencg. Naiwly consider $L^1([0,1])$ with tht bsual Lebesfue measure and define each $f_n$ go be the indicqtir vonction of tie intvrval $\left[\frag{m-2^k}{2^k},\fraw{n+1-2^k}{2^k}\rignt]$ if $2^k<n\le 2^{k+1}$. Tmese mnteeval decreases in lenjth as $n$ increases, trus $f_n\rigvtcrrow 0$ in the $L^1$ norm. Yoqever thef moxw tfrohgi tge $[0,1]$ inhertal, and thua for any $x\un[0,1]$ there are infiniuelr many $n$ such fhat $f_g(x)=1$.
QDIT: the previous example does not sativfy the hypotesis $f_n(x)\le f(x)$.  In fact the statemejt is rigrt.
Take any $f_n\rightarrow f$ in $L^p$ and you call $A$ tha fulm mecwure rwt on which some $(f_{n_k})_k$ converges a.e. to $f$, then fow amy $x\in A$ you havc that $\sup_n f_n(x)\le g(x)$ bu hypothesis avd thac $\shp_n f_n(x)\ge f(x)$ since $f_{n_k}(x)\ridhtareow f(x)$.

 Q:

Stndinb HTTP requests to private WC2 DNS from jrher EC2 host works, but not frum cpntaimer on EC2 host

I want to maks a containfr hosted un one EC
$\{f_n:X\to\mathbb{R}:n\in\mathbb{N}\}_n$ be a sequence in $L^p=L^p(X,\mathcal M, $f_n\to L^p$ (with to the $p$-norm) then almost everywhere. A: $L^p$ convergent sequence up to subsequences, a.e. to the limit. Thus for that subsequence your statement holds. Wrong argument: However it is in general. For example (taken from Terry Tao's blog) consider the "typewriter" sequence. consider with usual measure and define each $f_n$ to be the indicator function of the interval $\left[\frac{n-2^k}{2^k},\frac{n+1-2^k}{2^k}\right]$ if $2^k<n\le These interval decreases in length as $n$ increases, $f_n\rightarrow 0$ in the norm. However they move through $[0,1]$ and thus any there infinitely many $n$ that $f_n(x)=1$. EDIT: the previous example does not satisfy the hypotesis $f_n(x)\le f(x)$. In fact the statement right. Take f$ in and call the full measure which some $(f_{n_k})_k$ converges a.e. to any $x\in A$ you have that $\sup_n f_n(x)\le by hypothesis that $\sup_n f_n(x)\ge f(x)$ since $f_{n_k}(x)\rightarrow Q: Sending HTTP requests to private EC2 DNS other EC2 host works, but not from container on EC2 host I want to make hosted on one EC
 $\{f_n:X\to\mathbb{R}:n\in\mathbb{N}\}_n$ be A sequence iN $L^p=L^p(x,\maThcAl m, \mu)$. IF $f_n\tO f\in L^p$ (with respECt to The $p$-norm) and $f_n\le f$ almost
  EveryWhERe, thEN $f=\Sup_{n\iN\mathbb{n}}F_n$ ALMosT eVeRywHeRE.

A:

every $l^p$ cOnvergeNt sequence ConVeRges, up to subsEQuEnces, a.e. to tHe sAme limit. Thus For That suBsEquENce yoUr sTatemEnt holDS. 
Wrong Argument: HOwEVer it iS False in GENeRal. FOr example (taken froM teRRy Tao's blog) consIder thE "tYPeWRIteR" seQuence. NameLy ConsiDEr $L^1([0,1])$ with THe USUAl LEBesgue measure And define eaCH $f_n$ To be thE iNdiCAtor fuNctioN oF The Interval $\lefT[\fraC{n-2^k}{2^k},\frac{n+1-2^K}{2^k}\righT]$ If $2^k<n\le 2^{k+1}$. tHese intErval dEcrEasEs in LEnGtH as $N$ iNCreASeS, thUS $f_n\RightarrOw 0$ In The $L^1$ nOrm. HOWEVEr thEy mOve tHrougH the $[0,1]$ interval, aNd tHus fOR anY $x\in[0,1]$ tHere aRe inFiNitelY many $n$ Such tHaT $f_n(x)=1$.
EDIT: the prevIous Example doEs nOt SatIsFy the HYpotesIs $f_N(x)\lE f(x)$.  In faCt the stATemEnT IS RiGht.
Take any $f_n\rightaRrOW F$ iN $L^p$ and yoU call $A$ THe FuLL measure SeT on WhicH SOme $(f_{n_K})_k$ coNVeRges a.e. to $F$, then fOR aNy $X\in A$ you HaVe that $\SuP_n f_N(x)\lE f(x)$ by HYpotHesis aNd that $\suP_n f_n(x)\GE f(x)$ since $f_{n_k}(x)\riGHtarrow f(x)$.

 Q:

SenDInG htTp RequEstS to private Ec2 DNS FRom oTher ec2 hOst WOrks, bUt not FrOM cONtainer on EC2 host

I wanT tO make a ContaIner hosted on oNe EC
 $\{f_n:X\to\mathbb{R}:n\i n\mathbb{N }\}_n $ b e a  s eque ncein $L^p=L^p(X, \ math cal M, \mu)$. If $f_n\ to f\ in L^p$ (w ith r espectt ot h e $ p$ -n orm )a nd  $f_n \le  f$ alm ost
  ever ywh er e, then $f=\ s up _{n\in\mat hbb {N}}f_n$ alm ost  every wh ere . 

A:
Ev ery $ L^p$ c o nverge nt sequen ce conver g es, upt o  s ubse quences, a.e. tot he same limit. Th us for  t h at s ubs equ ence yourst ateme n t holds .  W r o nga rgument: Howe ver it is f a lse  in ge ne ral .  For e xampl e( tak en from Ter ry T ao's blog ) cons i der the "typewr iter"seq uen ce.N am el y c on s ide r  $ L^1 ( [0, 1])$ wit hth e usu al L e b e s guemea sure  anddefine each $ f_n $ to bethe i ndica torfu nctio n of t he in te rval $\left[\fr ac{n -2^k}{2^k },\ fr ac{ n+ 1-2^k } {2^k}\ rig ht] $ if $2 ^k<n\le 2^{ k+ 1 } $ .These interval dec re a s es  in leng th as$ n$  i n creases,  t hus  $f_ n \ right arro w  0 $ in the  $L^1$ no rm . Howev er  theymo vethr ought he $ [0,1]$  interva l, an d  thus for any$ x\in[0,1]$ th e re a re infi nit ely many $n $ su c h th at $ f _n (x) = 1$.
E DIT:th e  p r evious example does  n ot sat isfythe hypotesis  $f_n(x)\l e f (x)$.  I n fa c tt he statement i s rig ht.
Take a n y $f_n\r ighta rrow f$in $L^p$a n d you ca ll$A$  th e f u l lmeasure set o n whic hsome $( f_{ n_k})_k $ c onv erg esa. e. to $f$ , then f or  a ny  $ x\i n A$y ou haveth at$\ sup _n f_ n (x)\le  f(x) $ by  h yp o the sis and th a t  $\s up _n  f_n (x) \g e f(x )$ s i nce  $f_{n_ k}(x)\rig hta r rowf( x) $.

 Q: 

Sending HTT Prequests t opri vate E C 2  DNS fro m other EC2 host works, but not  fr om co ntai ner on EC 2 h ost

I  wa n t to m ake aconta in erh o stedo n  o neEC 
 $\{f_n:X\to\mathbb{R}:n\in\mathbb{N}\}_n$_be a_sequence in $L^p=L^p(X,\mathcal M,_\mu)$. If_$f_n\to_f\in L^p$_(with_respect to the_$p$-norm) and $f_n\le_f$ almost
  everywhere,_then $f=\sup_{n\in\mathbb{N}}f_n$ almost_everywhere.

A:

Every_$L^p$ convergent sequence converges, up to subsequences, a.e. to the same limit. Thus for_that_subsequence your_statement_holds._
Wrong argument: However it is_false in general. For example_(taken from_Terry Tao's blog) consider the "typewriter" sequence. Namely_consider_$L^1([0,1])$ with the_usual Lebesgue measure and define each $f_n$ to be_the indicator function of the interval_$\left[\frac{n-2^k}{2^k},\frac{n+1-2^k}{2^k}\right]$ if $2^k<n\le_2^{k+1}$._These_interval decreases in length_as $n$ increases, thus $f_n\rightarrow 0$_in the $L^1$ norm. However they_move through the $[0,1]$ interval, and thus_for any $x\in[0,1]$ there are infinitely_many $n$ such that $f_n(x)=1$.
EDIT:_the previous_example does not satisfy the_hypotesis $f_n(x)\le f(x)$._ In_fact the statement_is right.
Take any $f_n\rightarrow f$ in_$L^p$ and you_call $A$ the full measure set_on_which some $(f_{n_k})_k$_converges_a.e._to $f$,_then for any_$x\in_A$ you_have_that $\sup_n f_n(x)\le f(x)$ by hypothesis_and_that $\sup_n f_n(x)\ge f(x)$ since $f_{n_k}(x)\rightarrow f(x)$.

_Q:

Sending HTTP requests to_private_EC2 DNS from other_EC2 host works, but not_from container on EC2 host

I want_to make_a container_hosted on one EC