Let $f : \mathbb { R } _ { + } \rightarrow \mathbb { R }$, be a piecewise continuous function, strictly positive and integrable. For all $x > 0$, we define $$g ( x ) = \frac { 1 } { x } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t \quad \text { and } \quad h ( x ) = \frac { 1 } { x } g ( x ) = \frac { 1 } { x ^ { 2 } } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t .$$ Determine the limit of $g ( x )$ as $x$ tends to $+ \infty$. Denoting by $\mathbb { 1 } _ { [ 0 , x ] }$ the indicator function of $[ 0 , x ]$, you may note that $g ( x ) = \int _ { 0 } ^ { + \infty } \frac { 1 } { x } t f ( t ) \mathbb { 1 } _ { [ 0 , x ] } ( t ) \mathrm { d } t$.
Let $f : \mathbb { R } _ { + } \rightarrow \mathbb { R }$, be a piecewise continuous function, strictly positive and integrable. For all $x > 0$, we define
$$g ( x ) = \frac { 1 } { x } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t \quad \text { and } \quad h ( x ) = \frac { 1 } { x } g ( x ) = \frac { 1 } { x ^ { 2 } } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t .$$
Determine the limit of $g ( x )$ as $x$ tends to $+ \infty$. Denoting by $\mathbb { 1 } _ { [ 0 , x ] }$ the indicator function of $[ 0 , x ]$, you may note that $g ( x ) = \int _ { 0 } ^ { + \infty } \frac { 1 } { x } t f ( t ) \mathbb { 1 } _ { [ 0 , x ] } ( t ) \mathrm { d } t$.