I have a question that says "suppose that f is an integrable function over [a,b] (not necessarily continuous), then for any positive number > 0, there is a continuous function g over [a,b], and f(b) = g(b), f(a) = g(a) and $$ \int _a^b\:\left|f\left(x\right)-g\left(x\right)\right|dx\:<\:$$
So, I know that by Riemann Condition for Integrability: A bounded functionfdefined on [a, b] is Riemann integrable on [a, b] if and only if for all ε > 0, there exists a partition P(ε) of[a, b] such that S(f;P(ε))−S(f;P(ε))< ε.
Since g is continuous on [a, b], then g is uniformly continuous on [a, b](Theorem 4-10 of Kirkwood). Let ε >0. Then by the uniform continuity off, there exists δ(ε)>0 such that if x, $y∈[a, b]$ and$|x−y|< δ(ε)$, then $$\left|g\left(x\right)−g\left(y\right)\right|<\frac{ε}{b-a}$$
Let $P={x_(0), x_(1), x_(2), . . . , x_(n)}$ be a partition of [a, b] with $‖P‖ < δ(ε)$. On $[x_(i−1), x_(i)]$,g assumes a maximum and a minimum (by the Extreme Value Theorem), say at $x'_{\left(i\right)}$ and $x"_{\left(i\right)}$ respectively. Thus
$S(f;P)−S(f;P) =n∑i=1(f(x'_(i)−f(x"_(i))∆xi<εb−an∑i=1∆xi=\frac{ε}{b-a}(b−a)=ε.$
Is that enough to make me say that there is a continuous function g for the given conditions? If not, what am I missing here? I'm a little confused.
Hint:
Using Darboux criteria one observes the if $f$ is Riemann integrable in $[a,b]$, then for any given $\varepsilon>0$, there are step functions $\ell(x)$ and $u(x)$ such that $\ell(x)\leq f(x)\leq u(x)$ such that $$\begin{align} \int^a_b(u(x)-f(x))\,dx&\leq \int(u(x)-\ell(x))\,dx <\varepsilon/2\\ \int^a_b(f(x)-\ell(x))\,dx&\leq \int(u(x)-\ell(x))\,dx <\varepsilon/2 \end{align} $$ This in a way says that step functions are dense in the space of Riemann inegrable functions.
It can be seen that one can find continuous functions $\phi$ and $\psi$ on $[a,b]$ such that $$\begin{align} \int^a_b|u(x)-\phi(x)|\,dx&<\varepsilon/2\\ \int^a_b|\psi(x)-\ell(x)|\,dx& <\varepsilon/2 \end{align} $$
From this, you get $$\begin{align} \int^b_a|\psi-f|&<\varepsilon\\ \int^b_a|\phi-f|&<\varepsilon. \end{align}$$