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)?


1 Answer 1


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.


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