Let $��$ be a finite-dimensional $k$-algebra with $k$ algebraically closed. Bongartz has recently shown that the existence of an indecomposable $��$-module of length $n > 1$ implies that also indecomposable $��$-modules of length $n-1$ exist. Using a slight modification of his arguments, we strengthen the assertion as follows: If there is an indecomposable module of length $n$, then there is also an accessible one. Here, the accessible modules are defined inductively, as follows: First, the simple modules are accessible. Second, a module of length $n \ge 2$ is accessible provided it is indecomposable and there is a submodule or a factor module of length $n-1$ which is accessible.