How to show that if
$g_{\epsilon}(x) = \sum_{m=1}^M = f(m/M)\chi_{[(m-1)/M,m/M)}(x)$
and
$x \in [(m-1)/M,m/M)$ then $g_{\epsilon}(x) = f(m/M)$ and not $g_{\epsilon}(x) = \sum_{m=1}^M f(m/M)$?
How to show that if
$g_{\epsilon}(x) = \sum_{m=1}^M = f(m/M)\chi_{[(m-1)/M,m/M)}(x)$
and
$x \in [(m-1)/M,m/M)$ then $g_{\epsilon}(x) = f(m/M)$ and not $g_{\epsilon}(x) = \sum_{m=1}^M f(m/M)$?