$g_n = max {min (f_n, g), -g} to f$
$begingroup$
I am currently self studying Mathematical analysis by M. Apostol.
I got stuck in trying to understand
$\$ Theorem 10.30 $ $Let ${f_n}$ be a sequence of functions in $L(I)$ which converges almost everywhere on $I$ to a limit function $f$. Assume that there is a nonnegative function g in $L(I)$ such that
$$|f(x)| < g(x) text{ a.e. on I}$$
Then $fin L(I)$.
$\$ Proof $\$ Define a new sequence of functions $g_n$ on $I$ as follows :
$$g_n = max {min (f_n, g), -g}$$
$dots$
Then $|g_n (x)|leq g(x)$ almost everywhere on $I$, and it is easy to verify that $g_n to f$ almost everywhere on $I$. Therefore, by the Lebesgue dominated convergence theorem, $f in L(I)$.
Can someone please highlight how $g_n to f$?
convergence lebesgue-integral sequence-of-function
$endgroup$
add a comment |
$begingroup$
I am currently self studying Mathematical analysis by M. Apostol.
I got stuck in trying to understand
$\$ Theorem 10.30 $ $Let ${f_n}$ be a sequence of functions in $L(I)$ which converges almost everywhere on $I$ to a limit function $f$. Assume that there is a nonnegative function g in $L(I)$ such that
$$|f(x)| < g(x) text{ a.e. on I}$$
Then $fin L(I)$.
$\$ Proof $\$ Define a new sequence of functions $g_n$ on $I$ as follows :
$$g_n = max {min (f_n, g), -g}$$
$dots$
Then $|g_n (x)|leq g(x)$ almost everywhere on $I$, and it is easy to verify that $g_n to f$ almost everywhere on $I$. Therefore, by the Lebesgue dominated convergence theorem, $f in L(I)$.
Can someone please highlight how $g_n to f$?
convergence lebesgue-integral sequence-of-function
$endgroup$
add a comment |
$begingroup$
I am currently self studying Mathematical analysis by M. Apostol.
I got stuck in trying to understand
$\$ Theorem 10.30 $ $Let ${f_n}$ be a sequence of functions in $L(I)$ which converges almost everywhere on $I$ to a limit function $f$. Assume that there is a nonnegative function g in $L(I)$ such that
$$|f(x)| < g(x) text{ a.e. on I}$$
Then $fin L(I)$.
$\$ Proof $\$ Define a new sequence of functions $g_n$ on $I$ as follows :
$$g_n = max {min (f_n, g), -g}$$
$dots$
Then $|g_n (x)|leq g(x)$ almost everywhere on $I$, and it is easy to verify that $g_n to f$ almost everywhere on $I$. Therefore, by the Lebesgue dominated convergence theorem, $f in L(I)$.
Can someone please highlight how $g_n to f$?
convergence lebesgue-integral sequence-of-function
$endgroup$
I am currently self studying Mathematical analysis by M. Apostol.
I got stuck in trying to understand
$\$ Theorem 10.30 $ $Let ${f_n}$ be a sequence of functions in $L(I)$ which converges almost everywhere on $I$ to a limit function $f$. Assume that there is a nonnegative function g in $L(I)$ such that
$$|f(x)| < g(x) text{ a.e. on I}$$
Then $fin L(I)$.
$\$ Proof $\$ Define a new sequence of functions $g_n$ on $I$ as follows :
$$g_n = max {min (f_n, g), -g}$$
$dots$
Then $|g_n (x)|leq g(x)$ almost everywhere on $I$, and it is easy to verify that $g_n to f$ almost everywhere on $I$. Therefore, by the Lebesgue dominated convergence theorem, $f in L(I)$.
Can someone please highlight how $g_n to f$?
convergence lebesgue-integral sequence-of-function
convergence lebesgue-integral sequence-of-function
asked Dec 29 '18 at 10:44
user397197
add a comment |
add a comment |
1 Answer
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$begingroup$
Let $Omega $ a set of measure $0$ s.t. $|f(x)|<g(x)$ for all $xnotin Omega $. Let $xnotin Omega $. Since $f_n(x)to f(x)$, there is $Nin mathbb N$ s.t. for all $ngeq N$, $$|f_n(x)|<g(x).$$
In particular, if $ngeq N$, $$g_n(x)=max{f_n(x),-g(x)}=f_n(x).$$
answered Dec 29 '18 at 10:56
SurbSurb
38.2k94475
$endgroup$
$begingroup$
Now I see... Thanks a lot
$endgroup$
– user397197
Dec 29 '18 at 11:03
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
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$begingroup$
Let $Omega $ a set of measure $0$ s.t. $|f(x)|<g(x)$ for all $xnotin Omega $. Let $xnotin Omega $. Since $f_n(x)to f(x)$, there is $Nin mathbb N$ s.t. for all $ngeq N$, $$|f_n(x)|<g(x).$$
In particular, if $ngeq N$, $$g_n(x)=max{f_n(x),-g(x)}=f_n(x).$$
answered Dec 29 '18 at 10:56
SurbSurb
38.2k94475
$endgroup$
$begingroup$
Now I see... Thanks a lot
$endgroup$
– user397197
Dec 29 '18 at 11:03
add a comment |
$begingroup$
Let $Omega $ a set of measure $0$ s.t. $|f(x)|<g(x)$ for all $xnotin Omega $. Let $xnotin Omega $. Since $f_n(x)to f(x)$, there is $Nin mathbb N$ s.t. for all $ngeq N$, $$|f_n(x)|<g(x).$$
In particular, if $ngeq N$, $$g_n(x)=max{f_n(x),-g(x)}=f_n(x).$$
answered Dec 29 '18 at 10:56
SurbSurb
38.2k94475
$endgroup$
$begingroup$
Now I see... Thanks a lot
$endgroup$
– user397197
Dec 29 '18 at 11:03
add a comment |
$begingroup$
Let $Omega $ a set of measure $0$ s.t. $|f(x)|<g(x)$ for all $xnotin Omega $. Let $xnotin Omega $. Since $f_n(x)to f(x)$, there is $Nin mathbb N$ s.t. for all $ngeq N$, $$|f_n(x)|<g(x).$$
In particular, if $ngeq N$, $$g_n(x)=max{f_n(x),-g(x)}=f_n(x).$$
answered Dec 29 '18 at 10:56
SurbSurb
38.2k94475
$endgroup$
Let $Omega $ a set of measure $0$ s.t. $|f(x)|<g(x)$ for all $xnotin Omega $. Let $xnotin Omega $. Since $f_n(x)to f(x)$, there is $Nin mathbb N$ s.t. for all $ngeq N$, $$|f_n(x)|<g(x).$$
In particular, if $ngeq N$, $$g_n(x)=max{f_n(x),-g(x)}=f_n(x).$$
answered Dec 29 '18 at 10:56
SurbSurb
38.2k94475
edited Dec 29 '18 at 11:06
answered Dec 29 '18 at 10:56
SurbSurb
38.2k94475
answered Dec 29 '18 at 10:56
SurbSurb
38.2k94475
answered Dec 29 '18 at 10:56
SurbSurb
38.2k94475
38.2k94475
$begingroup$
Now I see... Thanks a lot
$endgroup$
– user397197
Dec 29 '18 at 11:03
add a comment |
$begingroup$
Now I see... Thanks a lot
$endgroup$
– user397197
Dec 29 '18 at 11:03
$begingroup$
Now I see... Thanks a lot
$endgroup$
– user397197
Dec 29 '18 at 11:03
$begingroup$
Now I see... Thanks a lot
$endgroup$
– user397197
Dec 29 '18 at 11:03
add a comment |
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