Theorem:
If $f: [a,b] \to \Bbb R$ is Riemann-integrable, then $f$ is Lebesgue-integrable, and:
$$ \int_a^b f(x) dx = \int_{[a,b]} fd\lambda$$
Here is the proof: (assuming that we know that step functions are Lebesgue integrable and that their Riemann integral coincides with their Lebesgue one)
The Riemann-integrability of $f$ implies the existence of two sequences of step functions $(g_n) \uparrow$ and $(h_n) \downarrow$, such that $g_n \le f \le h_n$ for all $n$, and:
$\int_a^b (h_n(x) - g_n(x))dx \to 0$ $(\star)$
$\int_a^b f(x) dx = \lim \int_a^bg_n(x)dx$
It is easy to see that $(g_n)$ and $(h_n)$ converge pointwise to some $g$ and $h$ $\in L^1$ (respectively).
Letting $\gamma = |g_1| + |h_1|$, we get that $h_n - g_n \le \gamma$ for all $n$, and $\gamma \in L^1$ being the sum of two $L^1$ functions. Moreover, $h_n - g_n \to h - g$ is a sequence of $L^1$ functions converging to an $L^1$ function. Therefore, by LDCT, we get:
$$\int_{[a,b]} (h_n - g_n) d \lambda \to \int_{[a,b]} (h - g) d \lambda$$
Which implies that:
$$\int_{[a,b]} (h - g) d \lambda = 0$$
Hence $h - g = 0$ $\lambda-$a.e. since it is $\ge 0$, i.e. $h = g$ $\lambda-$a.e.
Now, $g \le f \le h$, hence $f = g$ $\lambda-$a.e.
But from this, we can only conclude that $f$ is measurable with respect to the Lebesgue $\sigma-$algebra on $[a,b]$, and not the Borel one.
Are we assuming the Lebesgue $\sigma-$algebra on $[a,b]$? If not, how can we prove that $f$ is measurable?
The rest of the proof becomes straightforward.
A Riemann-integrable function is not necessarily Borel, see A Riemann integrable function $f$ on a bounded interval $[a, b]$ is measurable with respect to the Borel measure on $[a,b]$?
So you are going to have to assume you are working with the Lebesgue $\sigma$-algebra if you want to get a true theorem.