# Proof that $f$ is Riemann integrable without Lebesgue criteria

Consider $f$ and let it be Riemann integrable on $[0,1]$ and Lebesgue measurable on $[0,1]$. Show that function is Lebesgue integrable (without Lebesgue-theorem).

I thought about considering lower and upper Daurboux sums $M(x)$ and $m(x)$ and we know that $m(x) \le f(x) \le M(x)$ , but I don't use measurability of $f$. Also we know that Riemann integrable function is bounded, maybe it will be enough for Lebesgue integrability?

Any hints?

• Reimann integrable (roughly speaking) implies Lebesgue measurable: math.stackexchange.com/questions/291020/… – parsiad Mar 25 '18 at 15:27
• @parsiad yes, I understand that. But in my question I don't want to use Lebesgue theorem (if function is properly Riemann integrable , then it's Lebesgue integrable) – openspace Mar 25 '18 at 15:31

Since $f$ is bounded, we may shift it so that $f\geq0$. Since $f$ is Riemann-integrable, there is a sequence of step functions $h^+_k,h^-_k:[0,1]\to\mathbb R$ so that $0\leq h^-_k\leq f\leq h^+_k$ and $$\lim_{k\to\infty}\int_0^1h^+_k = \lim_{k\to\infty}\int_0^1h^-_k .$$ These are integrals of step functions and can be defined as finite sums. The limit is the Riemann integral $R$ of $f$.
The Lebesgue integral $L$ of $f$ is the supremum of all $\int w$ for $0\leq w\leq f$ with $w$ simple. Since $w\leq f\leq h_k^+$, we know that $\int w\leq\int h_k^+$ for every $k$. But $\int h_k^+\to R$, so in fact $\int w\leq R$ and consequently $L\leq R$.
This shows that he Lebesgue integral is finite and therefore $f$ is Lebesgue integrable. To get the value of the Lebesgue integral, simply use the step function $h_k^-$ as a simple function in the definition to see that in fact $L=R$.