Dgdrxrt

Consider $f in L^2(dmu)$, $ f_n in L^2(dmu)$, $f_n(x) to f(x)$ a.e. and $int |f_n|^2 , dmu mathoplongrightarrowlimits_n to infty int |f|^2 , dmu$. Use Egorov's theorem to show that $f_n to f$ in $L^2(dmu)$.

Egorov's Theorem requires that $f$ is defined or supported on a set $E$ of finite measure, so that is implicitly given by the question. For any $epsilon$, by Egorov's Theorem, there exists a $A_epsilon subset E$ such that $f_n$ converges uniformly to $f$ in $A_epsilon$ and that $m(E-A_epsilon) < epsilon$. For large enough $n$, $|f_n - f| < epsilon$ for all $x in A_epsilon$.

beginalign*
limlimits_n to infty left( lVert f_n - f rVert_2 right)^2 &= limlimits_n to infty int |f_n - f|^2 , dmu \
&= limlimits_n to infty int_A_epsilon |f_n - f|^2 , dmu + limlimits_n to infty int_E - A_epsilon |f_n - f|^2 , dmu \
&le epsilon^2 cdot m(E) + epsilon cdot limlimits_n to infty max(|f_n - f|^2) \
endalign*

The left term is arbitrarily small. I am stuck on trying to show that the term on the right is arbitrarily small.

I'm not using the given that:

beginalign*
limlimits_n to infty int |f_n|^2 , dmu &= int |f|^2 , dmu \
endalign*

Which means that the $L^2$ norm of $f_n$ approaches that of $f$. I presume that I'm supposed to use that but I don't see how.

I asked this question several days ago, but I didn't get an answer, so I'm rewriting a cleaned up version here.

Consider $f_n, f in L^2(dmu)$, $f_n(x) to f(x)$ a.e. and $|f_n|_2 to |f|_2$. Use Egorov's theorem to show that $f_n to f$ in $L^2(dmu)$.

Hint: $|f_n-f|^2 = |f_n|^2 + |f|^2 - 2cdotmathrmRe(f_noverlinef)$. Since $f_nto f$ almost everywhere, $mathrmRe(f_noverlinef)to |f|^2$ almost everywhere. Use Egorov's theorem to show that for all large $n$,
$$
bigg|int mathrmRe(f_noverlinef),dmu - int|f|^2,dmubigg| < epsilon.
$$
Then use that $int |f_n|^2,dmu to int |f|^2,dmu$ to conclude that $f_nto f$ in $L^2$.