A question about Riemann Integral, the function is not integrable, but the limit of Riemann sum exists Let $F(x)$ be a differentiable function over $\mathbb{R}$ and $f(x)$ the derivative of $F(x)$. By the fundamental increment lemma, there exists a function $\varphi$  such that
$$\lim_{h\to0}\varphi(h)=0\quad\text{and}\quad F(a+h)-F(a)=f(a)h+\varphi(h)h$$
for sufficiently small but non-zero $h$. In fact, it suffices to define
$$\varphi(h)=\begin{cases}
  \frac{F(a+h)-F(a)}{h}-f(a),& h\neq 0,\\
  0,& h=0.
\end{cases}$$
Let
$$a = x_0 < x_1 < x_2 < \dots < x_n = b$$
be a partition of $[a,b]$ and $\|\pi\|=\max\{\Delta x_i=x_{i+1}-x_i\}$, then
$$F(x_{i+1})-F(x_i)=f(x_i)\Delta x_i+\varphi_i(\Delta x_i)\Delta x_i$$
with $\lim_{\Delta x_i\to 0}\varphi_i(\Delta x_i)=0$.
Hence
$$F(b)-F(a)=\sum_{i=0}^{n-1}(F(x_{i+1})-F(x_i))=\sum_{i=0}^{n-1}f(x_i)\Delta x_i+
\sum_{i=0}^{n-1}\varphi_i(\Delta x_i)\Delta x_i.$$
It is easy to verify that
$$\lim_{\|\pi\|\to0}\sum_{i=0}^{n-1}\varphi_i(\Delta x_i)\Delta x_i=0.$$
Hence we have that
$$\lim_{\|\pi\|\to0}\sum_{i=0}^{n-1}f(x_i)\Delta x_i=F(b)-F(a).$$
So my question is:Is it right that the above limit also exists even when $f(x)$ is not integrable?
 A: Since you are always using left endpoints, the limit could exist even if $ f $ is not Riemann-integrable.  One sometimes says that a function $ f $ is Cauchy-integrable on $ [ a , b ] $ if this limit exists, although the resulting notion of integral has some bad properties and is not really accepted as a kind of integral.  (Cauchy himself only proposed this definition for functions known to be continuous, which he then proved are Riemann-integrable.  In 1915, Gillespie proved that a function is Riemann-integrable if and only if it is bounded and Cauchy-integrable.  But you have no reason to believe that $ f $ must be continuous or even bounded.)  So you appear to have proved that the derivative of a differentiable function $ F $ must be Cauchy-integrable, in a way that satisfies the Fundamental Theorem of Calculus in that the value of the integral is $ F ( b ) - F ( a ) $.
That said, I don't believe that you have really proved this!  (Indeed, it is false; if $ F $ is Volterra's Function, then its derivative exists but is not Cauchy-integrable.)  There is a lot hidden in this claim that $$ \lim _ { \lVert \pi \rVert \to 0 } \sum _ { i = 0 } ^ { n - 1 } \varphi _ i ( \Delta x _ i ) \Delta x _ i = 0 \text . $$  As you move from one partition to another (as you must for $ \lVert \pi \rVert $ to approach $ 0 $), the points $ x _ i $ change and so the functions $ \varphi _ i $ change.  If you have a fixed function $ \varphi _ i $, then sure, $ \varphi _ i ( \Delta x _ i ) \Delta x _ i \to 0 $ as $ \Delta x _ i \to 0 $, and even quickly enough to overcome that there are more terms in the sum.  But in general, $ \varphi _ i ( \Delta x _ i ) $ could change in an uncontrolled way as $ \varphi _ i $ changes.
To look at this alleged limit more closely, you want to prove that for each $ \varepsilon > 0 $, for some $ \delta > 0 $, whenever $ \lVert \pi \rVert < \delta $, the absolute value of the sum is less than $ \varepsilon $.  And you want to prove this by choosing $ \delta $ so that, whenever $ \Delta x _ i < \delta $ (which you'll have since $ \Delta x _ i \leq \lVert \pi \rVert $), $ \lvert \varphi _ i ( \Delta x _ i ) \rvert < \varepsilon / ( b - a ) $.  Because then $$ \Bigg \lvert \sum _ { i = 0 } ^ { n - 1 } \varphi _ i ( \Delta x _ i ) \Delta x _ i \Bigg \rvert \leq \sum _ { i = 0 } ^ { n - 1 } \lvert \varphi _ i ( \Delta x _ i ) \rvert \Delta x _ i \leq \sum _ { i = 0 } ^ { n - 1 } \frac \varepsilon { b - a } \Delta x _ i = \frac \varepsilon { b - a } \sum _ { i = 0 } ^ { n - 1 } \Delta x _ i = \frac \varepsilon { b - a } ( b - a ) = \varepsilon \text . $$  But unfortunately you cannot pick $ \delta $ in this way for all partitions at once, only for a particular function $ \varphi _ i $ (which in turn is determined by the value of $ x _ i $).
One thing that you can prove by this argument is that the derivative of a uniformly differentiable function is Cauchy-integrable, satisfying the FTC.  To say that $ F $ is uniformly differentiable on an interval is to say that the same function $ \varphi $ can be used at every point in the interval; then you can drop the subscripts on $ \varphi $ and the argument goes through.
Another thing that this argument proves is that the derivative of a differentiable function is Cauchy–Henstock–Kurzweil-integrable, also satisfying the FTC.  This is a kind of integral that I just made up, a variation of the Henstock–Kurzweil integral that uses only left endpoints.  The Henstock–Kurzweil integral may be defined much like the Riemann integral, but the constant $ \delta > 0 $ such that $ \lVert \pi \rVert < \delta $ is replaced by a function $ \delta \colon [ a , b ] \to \{ d \mid d > 0 \} $, so that instead of requiring $ \Delta x _ i < \delta $, we require $ t _ i - x _ i < \delta ( t _ i ) $ and $ x _ { i + 1 } - t _ i < \delta ( t _ i ) $, where $ t _ i $ is the tag in the subinterval $ [ x _ i , x _ { i + 1 } ] $.  In the left-endpoint version, this becomes $ \Delta x _ i < \delta ( x _ i ) $, and so we may let $ \varphi _ i $ give us $ \delta ( x _ i ) $.
Neither of these results are new; in fact, the derivative of any uniformly differentiable function is Riemann-integrable (and in fact continuous), and the derivative of any differentiable function is Henstock–Kurzweil-integrable, with the Fundamental Theorem of Calculus applying in each case.  So you have essentially proved weaker left-endpoint-only versions of these theorems.  But actually, by applying the Fundamental Increment Lemma to the half-subintervals of an arbitrary tagged partition (so start by writing $ F ( b ) - F ( a ) $ as $ \sum _ i \big ( F ( t _ i ) - F ( x _ i ) \big ) + \sum _ i \big ( F ( x _ { i + 1 } ) - F ( t _ i ) \big ) $ instead of as just $ \sum _ i \big ( F ( x _ { i + 1 } ) - F ( x _ i ) \big ) $ and use both $ F ( x _ i ) - F ( t _ i ) = f ( t _ i ) ( x _ i - t _ i ) + \varphi _ i ( x _ i - t _ i ) ( x _ i - t _ i ) $ and $ F ( x _ { i + 1 } ) - F ( t _ i ) = f ( t _ i ) ( x _ { i + 1 } - t _ i ) + \varphi _ i ( x _ { i + 1 } - t _ i ) ( x _ { i + 1 } - t _ i ) $), you can prove the full theorems!
