Lebesgue integrability of continuous function in closed interval I'm trying to show that a continuous function $f$ in $[a,b]$ is Lebesgue integrable, using approximation through step functions.
It is pretty trivial to show using the connection between Riemann integral and Lebesgue integral. But I have to show it only using step functions approximation.
Using the definition of Lebesgue integral which uses step functions approximation seemed to be easy but I'm confused.
Thanks for any help
 A: Hint: Like for Riemann integral, use uniform continuity.

Edit: I add some details. Thanks to uniform continuity, for all $\epsilon>0$, you can find $\delta>0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$. Now, let $(I_n)$ be a finite partition of $[a,b]$ such that $I_n$ is an interval of diameter less that $\delta$; for $n \geq 1$, let $y_n = \sup\limits_{x \in I_n} f(x)$ and $x_n= \inf\limits_{x \in I_n} f(x)$. You can considerate: $\displaystyle g_{\epsilon}(x)= \sum\limits_n x_n 1_{I_n}$ and $\displaystyle h_{\epsilon}(x)= \sum\limits_n y_n 1_{I_n}$. Notice that $g_{\epsilon} \leq f \leq h_{\epsilon}$.
Now $\displaystyle \int_a^b h_{\epsilon}(x)dx$ dicreases when $\epsilon \to 0$ and is bounded below, so it converges to $\displaystyle \ell= \inf\limits_{\epsilon>0} \int_a^b h_{\epsilon}(x)dx$. Moreover, $\displaystyle \left| \int_a^b h_{\epsilon}(x)dx-\int_a^bg_{\epsilon}(x)dx \right| \leq \epsilon(b-a)$ and for any step function $s \leq f$, $\displaystyle \int_a^b s(x)dx \leq \int_a^b h_{\epsilon}(x)dx$, hence $\displaystyle \int_a^b s(x)dx \leq \ell$.
In fact, $\displaystyle \ell= \int_a^b f(x)dx$.
A: First of all the definition of Lebesgue integral for a bounded function uses approximations by Simple functions and not only step functions(Step functions are simple functions).
Well, here is a more general statement:-
Any bounded measurable function defined on a set $E$ of finite measure is integrable. In fact any bounded integrable function defined on a set of finite measure is measurable.
So as $f$ is continuous in $[a,b]$ it is bounded and all continuous functions are measurable , the claim follows.
Let $|f|<M$ for some positive real number $M$.
Then let $E_{k}=\{x\in [a,b]: \frac{M(k-1)}{n}<f(x)<\frac{Mk}{n}\}$
for $-n\leq k \leq n$.
Then $\cup_{k}E_{k}=[a,b]$ and they are disjoint.
Define $$\phi_{n}=\sum_{k=-n}^{n}\frac{Mk}{n}\mathbf{1}_{E_{k}}$$
and $$\psi_{n}=\sum_{k=-n}^{n}\frac{M(k-1)}{n}\mathbf{1}_{E_{k}}$$.
Then $\psi_{n}\leq f\leq \phi_{n}$
And $\int\phi_{n}=\sum_{k=-n}^{n}\frac{Mk}{n}\lambda(E_{k})$
$\int \psi_{n}=\sum_{k=-n}^{n}\frac{M(k-1)}{n}\lambda(E_{k})$
So $$\sum_{k=-n}^{n}\frac{Mk}{n}\lambda(E_{k})\geq\inf_{\phi\geq f}\int\phi$$
And $$\sum_{k=-n}^{n}\frac{M(k-1)}{n}\lambda(E_{k})\leq \sup_{\psi\leq f}\int\psi$$
All integrals are taken over $[a,b]$.
Where $\phi,\psi$ are simple functions.
So substracting the second inequality from the first gives:-
$$\inf_{\phi\geq f}\int\phi-\sup_{\psi\leq f}\int\psi\leq \sum_{k=-n}^{n}\frac{M}{n}\lambda (E_{k})=\frac{M\lambda([a,b])}{n}=\frac{M(b-a)}{n}\xrightarrow{n\to\infty} 0$$.
Thus $\inf_{\phi\geq f}\int\phi-\sup_{\psi\leq f}\int\psi=0$ and this gives integrability .
A: Since a continuous function on a closed interval is bounded, the estimate $\displaystyle \int_a^b |f(x)| \ dx \le |b-a|\sup_{a \le x \le b}|f(x)|$ shows that $f$ is absolutely integrable.
