# Completion (!) of proof of VI.3.4 (completeness of L1) in Lang's Real and Functional Analysis

Lang starts the proof that the $\mathcal{L}^1(\mu_X, E)$ - defined as the space of pointwise a.e. limits of Cauchy sequences of step maps (defined to vanish outside sets of finite measure) - is complete in the $L^1$-norm (functions $X \to E$, $E$ Banach) by using the fact that step functions are dense in $\mathcal{L}^1$. This is obviously a true fact but I don't see how it follows from the three facts he has already proven (no monotone or dominated convergence):

• VI.3.1: A Cauchy sequence of step maps has a subsequence which both converges both pointwise a.e. and converges uniformly outside a set of arbitrarily small positive measure.
• VI.3.2: If $(f_n)$, $(g_n)$ are Cauchy sequences of step maps converging pointwise a.e. to $f$, $g$, then $\lim\int{f_n}$, $\lim\int{g_n}$ exist and are equal, and $||f_n - g_n||_1 \to 0$. (Thus, $\int$ is well-defined.)
• VI.3.3: If $(f_n)$ is a Cauchy sequence of step maps and converges pointwise a.e. to $f$, then the same for $|f_n|$ and $|f|$. (Thus, $||\cdot||_1$ is well-defined.)

Essentially what we haven't proved is that Cauchy+p.w. a.e convergence of step maps implies $L^1$-convergence to the limit function ... is this an obvious fact that I'm missing (i.e., easier than proving later theorems directly and invoking them)?

We can demonstrate the following stronger result: If $f \in \mathscr L^1$ and if $(f_n)$ is an approximating sequence of step maps for $f$, then $f_n \to_{\mathscr L^1} f$, i.e. $\lVert f - f_n\rVert_1 \to 0$. This is certainly something we would expect if we want $\mathscr L^1$ and $L^1$ to basically be the same space.
To prove it, notice that $(f_m - f_n)_{m \in \mathbb N}$ is an approximating sequence to $f - f_n$. Thus by the definition of the $L^1$-seminorm on the previous page in Lang, $\lVert f - f_n \rVert_1 = \lim_{m \to \infty} \lVert f_m - f_n \rVert_1$. This goes to $0$ as $n \to \infty$ because $(f_n)$ is Cauchy.