# $L^1$-Cauchy sequence of step functions that converges a.e. also converges in $L^1$

Let $$X$$ be a measure space and $$E$$ be a Banach space. A step function from $$X$$ to $$E$$ is a measurable function ($$E$$ takes the Borel measure) with finite image and whose support has finite measure.

If $$(f_n)$$ is a Cauchy sequence of step functions that converge a.e. to a function $$f$$, then do they also converge to $$f$$ in the $$L^1$$ seminorm?

The context for this question is that Lang seems to make this leap in his proof of Theorem 3.4 on page 133 of Real and Functional Analysis, and I want to know if the proof is still correct.

• Found a duplicate question math.stackexchange.com/questions/2287211/… May 6 '19 at 23:37
• You have missed crucial hypothesis. There is not hope for this if you don't assume that $f \in L^{1}$. May 6 '19 at 23:53
• There is no hope even if you do assume $f \in L^1$. I’ve written up a counterexample in which $f \in L^1$. May 7 '19 at 21:42

Cauchy in what sense? If they are pointwise Cauchy the here is a counterexample. On the real line with Lebesgue measure let $$f_n= \sum\limits_{k=1}^{n} \frac 1 k I_{(k,k+1)}$$. Then $$f_n \to f=\sum\limits_{k=1}^{\infty} \frac 1 k I_{(k,k+1)}$$ but $$f_n$$ does not tend to $$f$$ in $$L^{1}$$ because $$\int |f_n-f|=\infty$$ for each $$n$$.
If $$(f_n)$$ is Cauchy in $$L^{1} (X,B)$$ then it converges to some $$g$$ in $$L^{1}(X,B)$$ by completeness of the latter and $$f=g$$ in view of the fact that some subsequence of $$\|f_n -g\| \to 0$$ almost everywhere.
As Kavi Rama Murthy points out, if the sequence is Cauchy in $$L^1$$, then it converges to $$f$$ in $$L^1$$. He has provided a counterexample in the case that the sequence $$\{ f_n \}$$ is merely pointwise Cauchy. In his example, the pointwise limit $$f$$ is not in $$L^1$$. However, it is possible to construct a counterexample in which the limit function is in $$L^1$$. One such example is given by setting $$f_n = \frac{1}{n}I_{[0,n]}$$, where $$I_{[0,n]}$$ is the indicator function for the interval $$[0,n]$$. The sequence $$f_n$$ is (uniformly) Cauchy because if $$n,m \geq N$$, then $$|f_n(x) - f_m(x)| \leq \frac{2}{N}$$ for all $$x$$. The sequence tends pointwise to the zero function but does not converge to $$0$$ in $$L^1$$ because $$\| f_n \|_1 = 1$$ for al $$n$$.