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In Riemann integral, we define $$\int_a^b f:=\sup\left\{\int_a^b \varphi \mid \varphi \leq f \text{ where }\varphi \text{ step function}\right\}=\inf\left\{\int_a^b \varphi \mid f\leq \varphi \text{ where $\varphi $ step function}\right\},$$

whereas in Lebesgue integral, we define (if $f\geq 0$)

$$\int_a^b f:=\sup\left\{\int_a^b \varphi \mid 0\leq \varphi \leq f \text{ where }\varphi \text{ simple function}\right\}.$$

Except the fact that in Lebesgue, this definition is restricted to non-negative function, why in Lebesgue integral $$\sup\left\{\int_a^b \varphi \mid 0\leq \varphi \leq f \text{ where }\varphi \text{ simple function}\right\}=\inf\left\{\int_a^b \varphi \mid f\leq \varphi \text{ where }\varphi \text{ simple function}\right\},$$ is not required?

bof
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Bruce
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  • the infimum will not work to define an integral of Lebesgue because a simple function is bounded by definition, however $f$ doesn't need to be bounded to be Lebesgue integrable. However in the integral of Riemann $f$ need to be bounded to be Riemann integrable, so we can use the infimum to define it integral. – Masacroso Jan 20 '21 at 10:17
  • @Masacroso: Thank you. So, if $f$ is bounded (in Lebesgue case), will the infimum and supremum coincident or still won't ? – Bruce Jan 20 '21 at 10:25
  • it will coincide as far $f$ will be a Lebesgue measurable function – Masacroso Jan 20 '21 at 10:27

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