48484

证明：由$f$可积的定义知，对任意的$\varepsilon  > 0$，存在$N$，使得\[\int_a^b {\left| {f\left( x \right) - {{\left[ f \right]}_N}\left( x \right)} \right|dx}  < \frac{\varepsilon }{3}\]对于函数${{{\left[ f \right]}_N}}$，由$\bf{Lusin定理}$知，对任给$\delta  = \frac{\varepsilon }{{3N + 1}} > 0$，存在闭集${E_\delta } \subset \left[ {a,b} \right]$，以及存在$\left[ {a,b} \right]$上连续函数$\varphi ：$$\left| {\varphi \left( x \right)} \right| \le N$，使得$m\left( {{E_\delta }} \right) < \delta $，且当$x \in \left[ {a,b} \right]\backslash {E_\delta }$时，$\varphi \left( x \right) = {\left[ f \right]_N}\left( x \right)$，于是\[\int_a^b {\left| {{{\left[ f \right]}_N}\left( x \right) - \varphi \left( x \right)} \right|dx}  = \int_{{E_\delta }} {\left| {{{\left[ f \right]}_N}\left( x \right) - \varphi \left( x \right)} \right|dx}  \le \frac{{2\varepsilon N}}{{3N + 1}} < \frac{{2\varepsilon }}{3}\]所以有\[\int_a^b {\left| {f\left( x \right) - \varphi \left( x \right)} \right|dx}  \le \int_a^b {\left| {f\left( x \right) - {{\left[ f \right]}_N}\left( x \right)} \right|dx}  + \int_a^b {\left| {{{\left[ f \right]}_N}\left( x \right) - \varphi \left( x \right)} \right|dx}  < \varepsilon \]