When is a Lebesgue integrable function a Riemann integrable function ?
And if we have $f\in \mathcal{L}^1([0,1],\lambda)$, does it implies that $f$ is Riemann integrable, and why ?
[Math] When is a Lebesgue integrable function a Riemann integrable function
integrationlebesgue-integralriemann-integration
Related Solutions
To compute the Lebesgue integral of the function $f:[0,5]\to\mathbb R$ directly, one can note that $f_n\leqslant f\leqslant g_n$ for every $n\geqslant1$, where the simple functions $f_n:[0,5]\to\mathbb R$ and $g_n:[0,5]\to\mathbb R$ are defined by $$ f_n(x)=n^{-1}\lfloor nx\rfloor,\qquad g_n(x)=n^{-1}\lfloor nx\rfloor+n^{-1}. $$ By definition of the Lebesgue integral of simple functions, $$ \int_{[0,5]} f_n\,\mathrm d\lambda=n^{-1}\sum_{k=0}^{5n-1}k\cdot\lambda([kn^{-1},(k+1)n^{-1})), $$ that is, $$ \int_{[0,5]} f_n\,\mathrm d\lambda=n^{-2}\sum_{k=0}^{5n-1}k=\frac{5(5n-1)}{2n}. $$ Likewise, $$ \int_{[0,5]} g_n\,\mathrm d\lambda=n^{-1}\sum_{k=0}^{5n-1}(k+1)\cdot\lambda([kn^{-1},(k+1)n^{-1}))=\frac{5(5n+1)}{2n}. $$ Since the integrals of $f_n$ and $g_n$ converge to the same limit, $f$ is Lebesgue integrable and $$ \int_{[0,5]} f\,\mathrm d\lambda=\lim_{n\to\infty}\frac{5(5n\pm1)}{2n}=\frac{25}2. $$
Your proof is valid for a bounded function defined on the closed, bounded interval $[a,b]$, despite the apparent simplicity. It also relies on the fact that the Riemann and Lebesgue integrals are the same for step functions, as mentioned by Tony Piccolo.
The other proof you mention is also valid but takes you on a more roundabout path because it strings together a number of results, each of which is not altogether trivial to prove.
A bounded, measurable function defined on a set of finite measure is Lebesgue integrable.
If a sequence of measurable functions converges almost everywhere to $f$, then the limit function $f$ is measurable.
If $f$ is Riemann integrable on $[a,b]$, then there exists a sequence of simple (measurable) functions converging almost everywhere to $f$.
Adding even more complexity, the proof of the third statement that I know also uses the fact that the set of discontinuities of a Riemann integrable function must be of measure zero. It begins by constructing a partition of $[a,b]$ with dyadic intervals:
$$I_{n,k} = \begin{cases}[a + (b-a)\frac{k-1}{2^n}, a + (b-a)\frac{k}{2^n})\,\,\, k = 1 , \ldots, 2^n -1 \\ [a + (b-a)\frac{2^n-1}{2^n}, b] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, k = 2^n\end{cases} $$
With $m_k = \inf_{I_{n,k}}f(x)$, we can construct the sequence of simple functions
$$\phi_n(x) = \sum_{k=1}^{2^n} m_k \, \chi_{I_{n,k}}(x)$$
The sequence is increasing and, since $f$ is bounded, by monotonicity is convergent to some function $\phi$ such that $\phi_n(x) \uparrow \phi(x) \leqslant f(x).$ If $x \in I_{n,k}$ then the oscillation of $f$ over that interval satisfies
$$\sup_{u,v \in I_{n,k}}|f(u) - f(v)| \geqslant f(x) - \phi_n(x) \geqslant f(x) - \phi(x).$$
Thus, $f(x) \neq \phi(x)$ only at points where $f$ is not continuous, which must belong to a measure zero set if $f$ is Riemann integrable. Therefore, $f$ is almost everywhere the limit of a sequence of simple functions and is measurable.
I would agree that the second approach is "pretty complicated". However, I think it is not uncommon to find theorems with a variety of proofs ranging in complexity.
Best Answer
No, it doesn't.
It was proven by Lebesgue that a bounded function $f$ is Riemann integrable precisely when the set of points where $f$ is discontinuous has measure zero.