Why does Lang need finite measure support in the proof Lemma 3.2 (Real and Functional Analysis)?

Question

If anyone has the book, Lemma 3.2 and its proof are on pages 130-132.

Here is the statement of the lemma:

Let $\{g_n\}$ and $\{h_n\}$ be Cauchy sequences of step mappings of $X$ into $E$, converging almost everywhere to the same map. Then the following limits exist and are equal: $$\lim\int_X g_n=\lim\int_X h_n$$ Furthermore, the Cauchy sequences $\{g_n\}$ and $\{h_n\}$ are equivalent, i.e. $\{g_n-h_n\}$ is an $L^1$-null sequence.

For context, $X$ is a measure space, $E$ is a Banach space, and Lang has previously defined a step mapping from $X$ to $E$ as a function that is measurable, has finite image, and whose support has finite measure (this is not Lang's precise definition but an equivalent one).

The proof starts by showing that $\lim\int_X g_n$ (and $\lim\int_X h_n$ exist; once we have done that we can define $f_n=g_n-h_n$, and it only remains to show that $\int_X f_n$ and $\int_X |f_n|$ go to 0, from our knowledge that $f_n\rightarrow 0$ almost everywhere. It suffices to prove the latter.

Fixing $\epsilon>0$ and choosing $N$ such that the Cauchy condition for $f_n$ holds, we have to show that $\int_X|f_n|$ is arbitrarily small for $n$ arbitrarily great. To do this, Lang divides $X$ into three parts and bounds the integral on each part. This is the step that my question is about, because it seems sufficient to just divide $X$ into two parts.

First Lang lets $A\subseteq X$ be a set of finite measure which includes the support of $f_N$. Then $$\int_{X-A}|f_n|=\int_{X-A}|f_n-f_N|\leq\int_X|f_n-f_N|<\epsilon$$ as long as $n\geq N$.

Then Lang cuts $A$ into two parts, using Lemma 3.1, which I will reproduce the statement of:

Let $\{f_n\}$ be a Cauchy sequence of step mappings. Then there exists a subsequence which converges pointwise almost everywhere, and satisfies the additional property: given $\epsilon$ there exists a set $Z$ of measure $<\epsilon$ such that this subsequence converges absolutely and uniformly outside $Z$.

Lang uses this lemma to produce a set $Z\subseteq A$ with measure $< \frac{\epsilon}{1+\|f_N\|}$ (where $\|\cdot\|$ is sup norm) such that $f_n$ has a subsequence which converges absolutely on $A-Z$. Choosing $Z$ to also include the measure zero set on which $f_n$ doesn't converge to $0$, we see that this subsequence converges absolutely on $A-Z$ to $0$. Then clearly $\int_{A-Z} |f_n|<\epsilon$ for $n$ great enough in this subsequence, and we are left with the third part, $Z$.

For this, we do $$\int_Z|f_n|\leq \int_Z|f_n-f_N|+\int_Z|f_N|\leq \int_X|f_n-f_N|+\mu(Z)\|f_N\|<2\epsilon$$ which works for $n$ great enough in the subsequence.

A possible objection is that we have not proved that $\int_X|f_n|$ is eventually $<\epsilon$, only that it is frequently $<\epsilon$. However, the Cauchy condition will show the former from the latter.

My question is, what is the point of $A$? Why don't we apply Lemma 3.1 directly to $X$ and break $X$ up into $Z$ and $X-Z$? Apparently Lang would rather work in $A$ because it has finite measure, but I am not sure where he uses this condition in the proof.

Answer 1

I just figured out the answer. It really helped to write it all out. To show that $\int_{A-Z}|f_n|<\epsilon$ from the uniform convergence of the subsequence of $f_n$ on $A-Z$, we need the fact that $A-Z$ has finite measure. Uniform convergence says that we can make the sup norm, not the $L^1$ norm, of $f_n$, as small as we want. So we make it smaller than $\frac{\epsilon}{\mu(A)}$; then $\int_{A-Z}|f_n|<\epsilon$.

Answer 2

For my own understanding I will try to rewrite the proof. In this proof there are three norms at play :

1. The Banach space norm $|\cdot|$ defined for vectors in $E$.

2. The step map seminorm $||\cdot||_1$ defined for step maps $f:X\to E$ by $||f||_1:=\int |f|$

3. The sup norm $||\cdot||$ defined for any map $f:X\to E$ by $||f||:=\sup \{ |f(x)|:x\in X\}$

Now we begin the proof. Let $(g_n),(h_n)$ be Cauchy sequences of step maps with respect to $||\cdot||_1$, both converging almost everywhere to some $f:X\to E$. Using the properties of the integral for step maps we have

$$\bigg|\int g_n -\int g_m\bigg|=\bigg|\int ( g_n-g_m)\bigg|\leq\int |g_n-g_m|=||g_n- g_m||_1\to 0$$

as $n,m\to \infty$. This shows that that $\int g_n$ is a Cauchy sequence in $E$ with respect to $|\cdot|$. Since $E$ is complete, $\lim_{n\to\infty} \int g_n$ exist in $E$. The same argument shows that $\lim_{n\to\infty}\int h_n$ also exists in $E$.

Let $f_n=g_n-h_n$ for each $n$. Note that $(f_n)$ is a sequence of step maps which is Cauchy with respect to $||\cdot||_1$ and converges almost everywhere to zero. What remains to be shown is that $\int f_n$ and $\int |f_n|$ both converge to zero, for then $\lim_{n\to\infty} \int g_n=\lim_{n\to\infty} \int h_n$ and $ \lim_{n\to\infty} ||g_n-h_n||_1=0$. Moreover, from the inequality $|\int f_n|\leq \int |f_n|$ it suffices to show $\lim_{n\to\infty}\int |f_n|=0$.

Let $\varepsilon>0$ be given. There exist $N\geq 1 $ such that $m,n\geq N$ implies $||f_n-f_m||_1<\varepsilon$. Since $f_N$ is a step map, there exist a measurable set $A$ of finite measure outside of which $f_N$ vanishes. Then for all $n\geq N$ we have

$$\int_{X\setminus A} |f_n|=\int_{X\setminus A} |f_n-f_N|\leq \int|f_n-f_N|=||f_n-f_N||_1<\epsilon $$

By Lemma 3.1 there exist a measurable set $B$ with $\mu(B)<\frac{\varepsilon}{1+||f_N||}$ and a subsequece $(f_{n_k})$ of $(f_n)$ such that $(f_{n_k})$ converges uniformly on $X\setminus B$. Moreover, there exist a null set $N$ outside of which $(f_n)$ converges to zero pointwise. Let $Z:=B\cup N$. Then it follows that $(f_{n_k})$ converges uniformly to zero on $X\setminus Z$, and in particular on $A\setminus Z$. Hence there exist $N'\geq 1$ such that $k\geq N'$ implies $|f_{n_k}(x)|<\frac{\varepsilon}{1+\mu(A\setminus Z)}$ for all $x\in A\setminus Z$. Then for all $k\geq N'$ we have

$$ \int_{A\setminus Z} |f_{n_k}|\leq \int_{A\setminus Z} \frac{\varepsilon}{1+\mu(A\setminus Z)}= \frac{\varepsilon \mu(A\setminus Z)}{1+\mu(A\setminus Z)}<\varepsilon$$

If in addition $k$ large enough that $n_k\geq N$, then by the choice of $B$ we also have

$$ \int_Z |f_{n_k}|\leq \int_Z |f_{n_k}-f_N|+\int_Z |f_N|\leq ||f_{n_k}-f_N||_1 +\mu(Z)||f_N||<\varepsilon+\varepsilon=2\varepsilon$$

Putting everything together we obtain that, for $k$ large enough,

$$ ||f_{n_k}||_1= \int |f_{n_k}|=\int_{X\setminus A} |f_{n_k}|+\int_{A\setminus Z} |f_{n_k}|+\int_{A\cap Z} |f_{n_k}|$$

$$\leq \int_{X\setminus A} |f_{n_k}|+\int_{A\setminus Z} |f_{n_k}|+\int_{ Z} |f_{n_k}|$$

$$\leq \varepsilon + \varepsilon + 2\varepsilon = 4\varepsilon $$

As $\varepsilon>0$ is arbitrary, it follows that $||f_{n_k}||_1\to 0$ as $k\to \infty$.

We have shown that, for every $\varepsilon>0$, $||f_{n}||_1<\varepsilon$ for infinitely many $n$. The final step is to show that $||f_{n}||_1\to 0$ as $n\to \infty$. Let $\varepsilon>0$ be given. Since $(f_n)$ is Cauchy with respect to $||\cdot||_1$ there exist $N\geq 1$ such that $||f_n-f_m||_1<\varepsilon/2$ for all $n,m\geq N$. Also, by what we just did there exist $m\geq N$ such that $||f_{m}||_1<\varepsilon/2$. It follows that for all $n\geq N$ we have

$$ ||f_{n}||_1\leq ||f_n-f_m||_1+||f_m||_1<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$$

As $\varepsilon>0$ is arbitrary it follows that $||f_{n}||_1\to 0$ as $n\to \infty$.