I'm trying to show that $\forall f:[a,b]\to\mathbb R$ continuous then $ \int_{[a,b]}fd\lambda=\int_a^bf(x)dx$ where $\lambda$ is Lebesgue measure.
I've shown Riemann and Lebesgue Intergrals of bounded functions on a closed and bounded interval are same but I couldn't show it for every continuous functions.($f$ is not non-negative, it is only continuous)
My definitions :
- $$\int_{[a,b]}fd\lambda = \int_{[a,b]}f^+d\lambda-\int_{[a,b]}f^-d\lambda $$
where $f^+=\max\{f,0\}$ and $f^-=-\min\{f,0\}$ and $f=f^++f^-$ is measurable.
- $$\int fd\lambda = \sup\left\{\int \varphi d\lambda : \varphi \; \;\textrm{is simple and} \;\; \varphi \leq f\right\} $$
where $f$ is non-negative and measurable.
- $$\int_a^b f(x)dx = \sup\left\{\int_a^b \varphi(x) dx : \varphi \; \;\textrm{is step function and} \;\; \varphi \leq f\right\}=\inf\left\{\int^b_a \psi(x) dx : \psi \; \;\textrm{is step function and} \;\; \psi \geq f\right\} $$ (Riemann integral definition)
I have known that since $f^+$ is a non-negative function its Lebesgue integral on $[a,b]$ is : $\int_{[a,b]}f^+d\lambda=\int f^+.\chi_{[a,b]}d\lambda = \sup\left\{\int \varphi d\lambda : \varphi \; \;\textrm{is simple and} \;\; \varphi \leq f^+.\chi_{[a,b]}\right\}$ where $\chi_{[a,b]}$ is characteristic function of $[a,b]$.
In addition, I have shown that Riemann and Lebesgue integrals of step functions are equal on $[a,b]$.
How can I show $\int_{[a,b]}f^+d\lambda-\int_{[a,b]}f^-d\lambda\leq \int_a^b f(x)dx$ and $\int_{[a,b]}f^+d\lambda-\int_{[a,b]}f^-d\lambda\geq \int_a^b f(x)dx$ for every continuous functions?
I appreciate for any help
We know that if a function is Riemman integrable on a bounded interval,it is measurable,and Lebesgue integrable and their integrals coincide.
Here,since $f$ is continuous at a closed and bounded intervals,it is Riemann integrable.
If you have the result for bounded functions,then note that a continuous function on a bounded closed interval,is bounded.
So you do not have to involve simple function e.t.c