 from this apparent paradox? How then?

A:

That depends on what you mean by $\lim_{n\to \infty} g_n(x)$. The pointwise limit is zero, so the integral is zero. And if you put the parentheses so: $\lim_{n\to \infty} (g_n(x))$ then we have to conclude that the limit is zero and
$$\int_0^1 \lim_{n\to \infty} (g_n(x))\,dx=\int_0^1 0\,dx=0$$
However, you can also read this as $\lim_{n\to \infty} (g_n\, dx)$ where $dx$ denotes the usual Lebesgue measure. This is a limit of measures and it does, in fact, converge to the point measure at $0$, so that 
$$\int_{[0,1]} \lim_{n\to \infty} (g_n\, dx)=\int_{[0,1]}d\delta_0=1$$
(I've changed the notation because $\int_0^1$ can be confusing in this case, as $\int_{(0,1)}d\delta_0=0$).
So the "paradox" actually comes from different kinds of limits, and the arising confusion comes from not explaining the kind of convergence that is used in each case.

A:

The equality $\lim_{n\to\infty}g_n(x)=\delta(x)$ is incorrect. In fact, $\lim_{n\to\infty}g_n(x)=0$ for all $x$ (including $x=0$). For $D_n$, 
$$
\lim_{n\to\infty}D_n(x)=\begin{cases}0,&\ x\ne0,\\ \infty,&\ x=0\end{cases}
$$
So, saying that the two limits are equal makes no sense. 
The physicists can "get away with it" because they formally treat $\delta$ as a function, although it is not. This is well-understood by mathematicians. One uses the functions $g_n$, $D_n$ to define linear functionals over the space of continuous functions. Given a continuous function $f$, one defines
$$
\varphi_n:f\mapsto\int_0^1f
from this apparent paradox? How then? 

 A: 

 That depends on what you mean by $ \lim_{n\to \infty } g_n(x)$. The pointwise limit is zero, so the built-in is zero. And if you place the parentheses so: $ \lim_{n\to \infty } (g_n(x))$ then we have to conclude that the terminus ad quem is zero and 
 $ $ \int_0 ^ 1 \lim_{n\to \infty } (g_n(x))\,dx=\int_0 ^ 1 0\,dx=0$$ 
 However, you can also learn this as $ \lim_{n\to \infty } (g_n\, dx)$ where $ dx$ denotes the common Lebesgue meter. This is a limit of measure and it does, in fact, converge to the point measure at $ 0 $, so that 
 $ $ \int_{[0,1 ] } \lim_{n\to \infty } (g_n\, dx)=\int_{[0,1]}d\delta_0=1$$ 
 (I've change the notation because $ \int_0 ^ 1 $ can be confusing in this case, as $ \int_{(0,1)}d\delta_0=0 $). 
 So the " paradox " actually comes from different kind of limits, and the arising confusion comes from not explain the kind of convergence that is used in each font. 

 A: 

 The equality $ \lim_{n\to\infty}g_n(x)=\delta(x)$ is incorrect. In fact, $ \lim_{n\to\infty}g_n(x)=0 $ for all $ x$ (include $ x=0 $). For $ D_n$, 
 $ $ 
 \lim_{n\to\infty}D_n(x)=\begin{cases}0,&\ x\ne0,\\ \infty,&\ x=0\end{cases } 
 $ $ 
 So, saying that the two limits are equal makes no common sense. 
 The physicists can " get away with it " because they formally treat $ \delta$ as a function, although it is not. This is well - understood by mathematicians. One use the function $ g_n$, $ D_n$ to define linear functionals over the space of continuous functions. Given a continuous function $ f$, one specify 
 $ $ 
 \varphi_n: f\mapsto\int_0 ^ 1f
 frlm this apparent paradox? How then?

A:

That bwpends on whzt you mdan by $\lim_{n\to \infty} g_n(x)$. The 'oinrwise limit is zero, so the kntegral ps zero. Abd ih you put the pacsntheses so: $\ljl_{n\to \mnfty} (g_n(x))$ then ee have to conclude that tfe limit is zero and
$$\int_0^1 \lim_{n\to \infty} (d_n(x))\,dx=\iny_0^1 0\,fx=0$$
However, you san sjso dvab this as $\lim_{n\to \infty} (g_n\, dx)$ whsre $dx$ venotes the ususl Lebesgue measure. This id a pimit of measures wnd it does, in sqct, converge to the poput measure zt $0$, so that 
$$\int_{[0,1]} \lim_{n\to \infty} (g_n\, dx)=\inc_{[0,1]}d\delta_0=1$$
(I've xhqngfg the notatmon bebause $\int_0^1$ can be confgsing im this case, as $\inv_{(0,1)}d\deota_0=0$).
So the "paradox" actnally comes from difserent kitda of limits, and tye arishng wonfjwiov ckmxs rrom nlt xxplaining fhe kind of convergence that ix twed in each czse.

A:

Thq qquality $\lim_{n\to\infty}g_n(x)=\delta(x)$ is incorrtct. Ih fact, $\lim_{n\to\infty}g_n(x)=0$ fir all $x$ (including $x=0$). Vor $D_n$, 
$$
\lii_{n\to\infty}D_n(x)=\begin{cases}0,&\ x\ne0,\\ \infty,&\ x=0\end{cases}
$$
So, sayitg thet tht bru llmits are equal makes no sense. 
The physicists dam "net away with it" because thry fptmally treat $\dglta$ as a runction, although lt is njt. Thus is welj-undrrstood by mathematicians. Obe uses the yunxtions $g_n$, $D_n$ to deyine linear yunctipnals over the space of contnnuous functions. Hiven a ckvtinuous functiov $f$, ote defintr
$$
\varphi_n:f\mapsto\ine_0^1f
from this apparent paradox? How then? A: on you mean $\lim_{n\to \infty} g_n(x)$. so integral is zero. if you put parentheses so: $\lim_{n\to \infty} (g_n(x))$ then have to conclude that the limit is zero and $$\int_0^1 \lim_{n\to \infty} (g_n(x))\,dx=\int_0^1 However, you can also read this as $\lim_{n\to \infty} (g_n\, dx)$ where $dx$ the Lebesgue This a limit of measures and it does, in fact, converge to the point measure at $0$, that $$\int_{[0,1]} \lim_{n\to \infty} (g_n\, dx)=\int_{[0,1]}d\delta_0=1$$ (I've changed notation because $\int_0^1$ can confusing in this case, as So "paradox" actually from kinds limits, and the confusion comes from not explaining the kind of convergence that is used in each case. A: The $\lim_{n\to\infty}g_n(x)=\delta(x)$ is fact, $\lim_{n\to\infty}g_n(x)=0$ all (including For $D_n$, $$ \infty,&\ x=0\end{cases} $$ So, saying that are equal makes no sense. The physicists can away with because they formally treat $\delta$ as function, although it is not. This is well-understood mathematicians. One uses the functions $g_n$, $D_n$ to define linear functionals over the space of Given a continuous function one defines $$
 from this apparent paradox? HoW then?

A:

That DepenDs oN whAt You mEan bY $\lim_{n\to \infty} g_n(X)$. the pOintwise limit is zero, so tHe intEgRAl is ZErO. And iF you put THe PARenThEsEs sO: $\lIM_{n\To \infTy} (g_N(x))$ then wE have to conCluDe That the limit IS zEro and
$$\int_0^1 \lIm_{n\To \infty} (g_n(x))\,dx=\Int_0^1 0\,Dx=0$$
HoweVeR, yoU Can alSo rEad thIs as $\liM_{N\to \infTy} (g_n\, dx)$ wheRe $DX$ denotES the usuAL leBesgUe measure. This is a lIMiT Of measures and iT does, iN fACt, CONveRge To the point MeAsure AT $0$, so that 
$$\INt_{[0,1]} \LIM_{N\to \INfty} (g_n\, dx)=\int_{[0,1]}d\dElta_0=1$$
(I've chanGEd tHe notaTiOn bECause $\iNt_0^1$ can Be COnfUsing in this Case, As $\int_{(0,1)}d\delTa_0=0$).
So thE "Paradox" ACtually Comes fRom DifFereNT kInDs oF lIMitS, AnD thE AriSing confUsIoN comeS froM NOT ExplAinIng tHe kinD of convergencE thAt is USed In eacH case.

a:

The EqUalitY $\lim_{n\tO\inftY}g_N(x)=\delta(x)$ is incorRect. in fact, $\lim_{N\to\InFty}G_n(X)=0$ for aLL $x$ (inclUdiNg $x=0$). for $D_n$, 
$$
\liM_{n\to\infTY}D_n(X)=\bEGIN{cAses}0,&\ x\ne0,\\ \infty,&\ x=0\end{caSeS}
$$
sO, sAying thaT the twO LiMiTS are equaL mAkeS no sENSe. 
The PhysICiSts can "geT away wITh It" Because ThEy formAlLy tReaT $\deltA$ As a fUnctioN, althougH it is NOt. This is well-unDErstood by mathEMaTICiANs. OnE usEs the functiOns $g_N$, $d_n$ to DefiNE lIneAR funcTionaLs OVeR The space of continuouS fUnctioNs. GivEn a continuous Function $f$, oNE DEfines
$$
\vaRphi_N:F\mAPsto\int_0^1f
 from this apparent parado x? How the n?

A :
Tha tdepe ndson what you me a n by  $\lim_{n\to \infty} g _n(x) $. Thep oi ntwis e limit is z ero ,so  th ei nt egral  is  zero.And if you  pu tthe parenthe s es  so: $\lim _{n \to \infty}(g_ n(x))$  t hen we ha veto co nclude that t he limitis zero a n d
$$\in t _ 0^ 1 \l im_{n\to \infty}( g_ n (x))\,dx=\int_ 0^1 0\ ,d x =0 $ $ 
Ho wev er, you ca nalsor ead thi s  a s $ \li m _{n\to \infty } (g_n\, dx ) $ w here $ dx $ d e notesthe u su a l L ebesgue mea sure . This is  a lim i t of me a sures a nd itdoe s,in f a ct ,con ve r get othe poi nt measu re  a t $0$ , so t h a t 
$ $\i nt_{ [0,1] } \lim_{n\to\in fty} (g_ n\, d x)=\i nt_{ [0 ,1]}d \delta _0=1$ $(I've changed t he n otation b eca us e $ \i nt_0^ 1 $ canbecon fusingin this cas e, a s  $ \int_{(0,1)}d\delt a_ 0 = 0$ ).
So th e "par a do x" actually  c ome s fr o m  diff eren t  k inds oflimits ,  a nd  the ar is ing co nf usi oncomes from  not e xplainin g the kind of conver g ence that isu se d in each  ca se.

A:

Th e eq u alit y $\ l im _{n \ to\in fty}g _n ( x) = \delta(x)$ is incor re ct. In  fact , $\lim_{n\to \infty}g_n ( x ) =0$ forall$ x$ (including $x= 0$).For $D_n$, 
$$
\lim _{n\t o\infty} D_n(x)=\b e g in{cases }0, &\x\n e0, \ \  \ infty,&\ x=0\ e n d{ca se s}
$$
S o,sayingtha t t hetwo  l imits are  equal m ak es  n osen se. T he physi ci sts  c an"geta way wi th it " be ca us e  th ey form a ll y trea t$\ delt a$as  a fu ncti o n,althoug h it is n ot. This  i swell-un derstood by m at hematician s.  On e uses t he funct ions $g_n$, $D_n$ to de f ine lin ear  func tion als overthe  space  of contin uous f uncti on s.G i ven a c on tin uo us functio n $f$ , one  d efin es
$$
\ varphi_n:f\mapsto\ i nt_ 0^1f
 from_this apparent_paradox? How then?

A:

That depends_on what_you_mean by_$\lim_{n\to_\infty} g_n(x)$. The_pointwise limit is_zero, so the integral_is zero. And_if_you put the parentheses so: $\lim_{n\to \infty} (g_n(x))$ then we have to conclude that_the_limit is_zero_and
$$\int_0^1_\lim_{n\to \infty} (g_n(x))\,dx=\int_0^1 0\,dx=0$$
However, you_can also read this as_$\lim_{n\to \infty}_(g_n\, dx)$ where $dx$ denotes the usual Lebesgue_measure._This is a_limit of measures and it does, in fact, converge_to the point measure at $0$,_so that 
$$\int_{[0,1]}_\lim_{n\to_\infty}_(g_n\, dx)=\int_{[0,1]}d\delta_0=1$$
(I've changed the_notation because $\int_0^1$ can be confusing_in this case, as $\int_{(0,1)}d\delta_0=0$).
So the_"paradox" actually comes from different kinds of_limits, and the arising confusion comes_from not explaining the kind_of convergence_that is used in each_case.

A:

The equality $\lim_{n\to\infty}g_n(x)=\delta(x)$_is incorrect._In fact, $\lim_{n\to\infty}g_n(x)=0$_for all $x$ (including $x=0$). For_$D_n$, 
$$
\lim_{n\to\infty}D_n(x)=\begin{cases}0,&\ x\ne0,\\_\infty,&\ x=0\end{cases}
$$
So, saying that the two_limits_are equal makes_no_sense._
The physicists_can "get away_with_it" because_they_formally treat $\delta$ as a function,_although_it is not. This is well-understood by_mathematicians. One uses the_functions_$g_n$, $D_n$ to define_linear functionals over the space_of continuous functions. Given a continuous_function $f$,_one defines
$$
\varphi_n:f\mapsto\int_0^1f