\( Check that if \)x \in Q-B\(, then \)f_{n j}(x)\( converges to a finite limit as
\)j \uparrow \infty\(. Hint: \)\quad x \in Q-B\( means that
\)\left|f_{n_{j+1}}(x)-f_{n j}(x)\right|<2^{-j / 3}\( for all sufficiently large
\)j\(.
By the Chebyshev inequality [Exercise 1.1.12],
$$
m\left(A_{j}\right) \leqslant 2^{2 j /
3}\left\|f_{n_{j+1}}-f_{n_{j}}\right\|^{2} \leqslant 2^{-j / 3},
$$
and since this is the general term of a convergent sum, the Borel-Cantelli
lemma, just before Exercise \)1.1 .11\( tells you that
$$
m(B)=m\left(A_{j} \quad \text { i.o. }\right)=0 .
$$
Therefore,
$$
C=\left(x: f_{n j}(x) \text { fails to converge as } j \uparrow \infty\right)
\subset B
$$
is of measure 0 , too [see Exercise 10], and you can define a measurable
function \)f\( by the rule
$$
f(x)= \begin{cases}0 & \text { if } x \in C \\ \lim _{j \uparrow \infty}
f_{n_{j}}(x) & \text { otherwise }\end{cases}
$$
The claim is that \)f_{n}\( converges to this function \)f\( in
\)\mathrm{L}^{2}(Q)\(. Because \)m(C)=0\(, \)f_{n}\(, converges to \)f\( pointwise
a.e., and by Fatou's lemma
$$
\begin{aligned}
\left\|f_{n}-f\right\|^{2} &=\int_{Q} \lim _{j \uparrow
\infty}\left|f_{n}-f_{n j}\right|^{2} \leqslant \liminf _{j \uparrow \infty}
\int_{Q}\left|f_{n}-f_{n j}\right|^{2} \\
&=\liminf _{j \uparrow \infty}\left\|f_{n}-f_{n j}\right\|^{2} \leqslant
2^{-i} \quad \text { if } \quad n \geqslant n_{i}
\end{aligned}
$$
From this you learn two things: first, that \)f \in \mathrm{L}^{2}(Q)\(, since
$$
\|f\| \leqslant\left\|f_{n_{1}}\right\|+\left\|f_{n_{1}}-f\right\|
\leqslant\left\|f_{n_{1}}\right\|+2^{-1 / 2}<\infty,
$$and second, that \)f_{n}\( approaches \)f\( in \)\mathrm{L}^{2}(Q)\( since \)2^{-i}
\downarrow 0\( as \)i \uparrow \infty\(. The proof is finished.
Step 4: Verify that \)\mathrm{L}^{2}(Q)\( is separable. This will finish the
proof that it is a Hilbert space. The problem is to show that you can come as
close as you please to any function \)f \in \mathrm{L}^{2}(Q)\( by functions
from a fixed countable family \)K \subset L^{2}(Q)\(. Clearly, it suffices to
deal with real functions. For any such function \)f\(, any integral \)i, j\(, and
\)k \geqslant 1\(, let
$$
\begin{aligned}
&f_{1}=\left\\{\begin{array}{lll}
f & \text { if } & |x| \leqslant i \\
0 & \text { if } & |x|>i
\end{array}\right. \\
&f_{2}= \begin{cases}f_{1} & \text { if }\left|f_{1}\right| \leqslant j \\
0 & \text { if }\left|f_{1}\right|>j\end{cases}
\end{aligned}
$$
and
$$
f_{3}=k^{-1}\left[k f_{2}\right],
$$
in which \)[f]\( stands for the largest integer \)\leqslant f\(. Each of these
functions is measurable and
$$
\begin{aligned}
\left\|f-f_{1}\right\|^{2} &=\int_{Q \cap(x ;|x|>i)}|f|^{2}, \\
\left\|f_{1}-f_{2}\right\|^{2} &=\int_{Q
\cap\left(x:\left|f_{1}\right|>j\right)}\left|f_{1}\right|^{2} \leqslant
\int_{Q \cap(x|| f \mid>j)}|f|^{2}, \\
\left\|f_{2}-f_{3}\right\|^{2} & \leqslant \int_{Q \cap(x:|x| \leqslant i)}
k^{-2} \leqslant 2 i k^{-2},
\end{aligned}
$$
so that you can make
$$
\left\|f-f_{3}\right\|
\leqslant\left\|f-f_{1}\right\|+\left\|f_{1}-f_{2}\right\|+\left\|f_{2}-f_{3}\right\|
$$
as small as you like by taking \)i, j\(, and \)k\( large, in that order. Bring in
the family \)K\( of piecewise constant functions \)f\(, which vanish far out, take
rational values only, and jump only at a finite number of rational points. \)K\(
is a countable subfamily of \)L^{2}(Q)\( and is closed under finite sums with
rational coefficients. Now \)f_{3}=l / k\( on the set
$$
A=Q \cap(x:|x| \leqslant i) \cap\left(x: l / k \leqslant f_{2}(x)<(l+1) /
k\right),
$$
so \)f_{3}\( is the sum (with coefficients \)l / k\( ) of indicator functions
\)1_{A}\( of sets of measure \)\leqslant 2 i<\infty\(. Therefore, to finish the
proof, you only have to check that such an indicator function can be well
approximated by a function from \)K\(. By the recipe for Lebesgue measure of
Section 1.1, you can cover such a set \)A\( by a countable family of intervals
\)I_{k}\(, so closely as to make
$$
0 \leqslant \sum_{k=1}^{\infty} m\left(I_{k}\right)-m(A)
$$
as small as you please. Especially, you can find a finite union
\)B=\bigcup_{k=1}^{n} I_{k}\( so as to make
$$
m(A-B)+m(B-A)=\int_{A-B} 1+\int_{B-A} 1=\left\|1_{A}-1_{B}\right\|^{2}
$$
as small as you like, and you can even modify \)B\( to make the endpoints of the
intervals rational, at the expense of a small change in
\)\left\|1_{A}-1_{B}\right\| .\( But then \)1_{B} \in K$, and the proof is
finished.
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