Lebesgue integral example

Question

please help me with the Lebesgue integration on the function

$$ f(x)= \begin{cases} 0 & \text{if $x$ < 0}\\ 1/2 & \text{if $x$ = 0}\\ 1 & \text{if $x$ > 0} \end{cases} $$

$$ \\\int_a^b{f}d\mu ? $$ can we take $\mu([a,b))$?

thank you!

Answer 1

I'm not sure what measure you are using, or how you define $\int_a^b f d \mu$ (ie, open, closed or half-open interval). (I am assuming that $\int_a^b f d\mu = \int_{[a,b]} f d \mu$, if not, adjust accordingly below.)

Note that $f$ is a simple function, $f = \frac{1}{2}1_{\{ 0 \}} + 1_{(0,\infty)}$.

Hence \begin{eqnarray} \int_{[a,b]} f d \mu &=& \int f 1_{[a,b]}d \mu \\ &=& \int (\frac{1}{2}1_{\{ 0 \}} + 1_{(0,\infty)}) 1_{[a,b]}d \mu \\ &=& \int (\frac{1}{2}1_{\{ 0 \} \cap [a,b]} + 1_{(0,\infty) \cap [a,b]} )d \mu \\ &=& \frac{1}{2}\mu ( \{ 0 \} \cap [a,b]) + 1 \mu( (0,\infty) \cap [a,b]) \end{eqnarray}

Answer 2

If $\mu$ is the Lebesgue measure, then $f=1_{(0,\infty)}$ $\mu$-almost everywhere and hence $$ \int_a^b f\,\mathrm d\mu=\int_a^b1_{(0,\infty)}\,\mathrm d\mu=\int_\mathbb{R}1_{(0,\infty)\cap \,[a,b)}\,\mathrm d\mu=\mu((0,\infty)\cap [a,b)). $$