Show $s(f) = \int_{\_ a}^b f $

Where $s(f) = sup \{ \int h d\lambda : h \in C[a,b], h \leq f\}$

And $\int_{\_ a}^b f $ is the Lower Darboux Integral

edit I know that $f: [a,b] \rightarrow R$ is bounded, but I do not have that it is continuous.

So far I have:

Let $\epsilon >0$, then $ \exists h \in C[a,b]$ $ s.t. h \leq f$ and $\int_a^b h d\lambda \geq s(f) - \epsilon$.

Since $f(x)\geq h(x)$, we have

$s(f)-\epsilon \leq \int_a^b h d\lambda = \int_{\_a}^b h \leq \int_{\_a}^b f$

I can't figure out how to show $\int_{\_a}^b f \leq s(f)$ though.


Ah, but that's the easy direction. Since $f\leq f$, $$\int_a^b f d\lambda\in \left\{\int_a^bhd\lambda:h\in C[a,b],h\leq f\right\}.$$ Hence $\int_a^bfd\lambda\leq s(f)$

Edit: For the updated question: Fix a partition $P$ on $[a,b]$ and let $s_{P,f}$ be the simple function defined by $$s_{P,f}=\sum_im_i\chi_{[x_i,x_{i-1}]}$$ with $m_i=\min\{f\vert_{[x_i,x_{i-1}]}\}$. For any $\epsilon>0$ we can find a continuous $h_{P,f}\leq s_{P,f}\leq f$ with $$\int_a^bs_{P,f}-h_{P,f}<\epsilon\Rightarrow \int_a^b s_{P,f}<\int_a^bh_{P,f}+\epsilon\;.$$ Hence, $$\underline{\int_a^bf}=\sup_P\left\{\int_a^bs_{P,f}\right\}\leq \sup_P\left\{\int_a^bh_{P,f}\right\}+\epsilon\leq s(f)+\epsilon$$

  • $\begingroup$ I should have specified: I know that f is bounded. Not continuous so it isn't an element of the set.... $\endgroup$ – Math Lady Oct 29 '18 at 13:56
  • $\begingroup$ @MathLady I've updated the answer. $\endgroup$ – user113529 Oct 30 '18 at 4:33

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