So, I'm trying to prove that if $f$ is a continuous function defined on a closed interval $[a,b]$ then it is Lebesgue integrable. Here is my attempt based on what I've found on the internet so far:
Since $[a,b]$ is compact, any continuous function $f$ will be uniformly continuous. Therefore, for any $\epsilon>0$ we can find a delta which is independent from the points $x$ and $y$ (due to uniform continuity) that satisfies this statement:
$$\forall \epsilon>0, \exists \delta>0 , \forall x,y \in [a,b] \text{ such that } |x-y|<\delta \implies |f(x)-f(y)|<\epsilon$$
Now we can partition the interval $[a,b]$ into $n$ subintervals $(I_n)$ such that each interval has a length less than $\delta$. For $n \in \mathbb{N}$ we define $\displaystyle a_n = \sup_{x \in I_n}f(x)$ and $\displaystyle b_n = \inf_{x \in I_n} f(x)$.
In other words, for any points in the subinterval $I_n$ we associate two numbers $a_n$ and $b_n$ whose values are determined by the supremum and infimum of $f$ on that subinterval respectively. It's obvious from the definition of $\sup$ and $\inf$ that in each subinterval we have $b_n \leq f \leq a_n$.
Now we define two simple functions that engulf the function $f$ from the above and the below by using $a_n$ and $b_n$ in the following way: $\displaystyle \varphi_{\delta} = \sum_{i=1}^n a_i. \mathbb{1}_{I_n}$ and $\displaystyle \psi_{\delta} = \sum_{i=1}^n b_i. \mathbb{1}_{I_n}$ Where $1_{I_n}$ denotes the indicator function on the subinterval $I_n$.
It is easy to see that $\psi_{\delta} \leq f \leq \varphi_{\delta}$ where the little subscript $_{\delta}$ denotes the dependence of the functions $\varphi_{\delta}$ and $\psi_{\delta}$ to $\delta$ by their definitions using $I_n$. Do not forget that the existence of $\delta$ itself depends on the choice of $\epsilon$. So, the value of $\delta$ is a function of $\epsilon$.
By definition of the Lebesgue integral for simple functions we know that:
$$ \int_{[a,b]} \varphi_{\delta} d\mu = \sum_{i=1}^n a_i. \mu(I_n) $$
$$ \int_{[a,b]} \psi_{\delta} d\mu = \sum_{i=1}^n b_i. \mu(I_n) $$
So far so good, the problem is that I don't know how I can relate these to the existence of the Lebesgue integral for $f$. In other words I have no idea what inequalities can guarantee the existence of $\int_{[a,b]} f d\mu$.
I'm using this definition: A function $f$ is Lebesgue integrable if and only if there exists a sequence of simple functions which are integrable and uniformly converge to $f$.
I'm not sure how this applies to my case. I'm afraid that I'm not able to show why $\varphi_{\delta}$ or $\psi_{\delta}$ converges uniformly to $f$.
Any ideas will be appreciated.
Choose $\varepsilon:=1/n$, (and choose $\delta=:\delta(1/n)$ for that) so that both $\varphi_n:=\varphi_{\delta(1/n)}$ and $\psi_n:=\psi_{\delta(1/n)}$ will uniformly converge to $f$, because of the construction (the proof is John's comment:) $$\|f-\varphi_{\delta(1/n)}\|_\max \ <\ 1/n$$