# Show $s(f) = \int_{\_a}^b f$

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, $$\underline{\int_a^bf}=\sup_P\left\{\int_a^bs_{P,f}\right\}\leq \sup_P\left\{\int_a^bh_{P,f}\right\}+\epsilon\leq s(f)+\epsilon$$