How to prove that if $int_L vec{F}(vec{x})cdot dvec{x}$ exists, then integral on separate curve exists?

I'll leave the proof that persuaded myself here, since I'm a beginner, anyone who would willing to look into it and find any mistake are very welcome.

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Part 1

Since $int_Lvec{F}(vec{x})cdot dvec{x}$ exists, we have the following conditions:

1. $vec{F}(vec{x})$ have definition on $L$ since $L$ is close set, $vec{F}(vec{x})$ is limited
2. $forallepsilon>0,existsdelta,foralllambda<delta$ for any partion $P$ and pick of points $vec{xi}_i$, we have $left|sum_{i=1}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Iright|<frac{epsilon_1}{2}$
3. $lim_{lambdato0}sum_{i=1}^m vec{F}(vec{x}_i)cdotDelta vec{x}_i=I$

Obviously there will include partitions that separates $L_1,L_2$ on $vec{x}_kin P$.

Therefore we have $left|sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i+sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Iright|<epsilon$

and also $lim_{lambdato0}sum_{i=1}^n vec{F}(vec{x}_i)cdotDelta vec{x}_i=lim_{lambdato0}left{sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i+sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_iright}=I$

Where we define the first part of the sum is from $L_1$ and the second part is from $L_2$

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Part 2

Now we assume that $lim_{lambdato0}sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i$ does not converge.

List the assumption as $existsepsilon_1>0,foralldelta>0,existslambda_1<delta$ exists partition and pick of points $exists(P^1,vec{xi}^1_i)$

That for all $Ainmathbb{R}$ we have $left|sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Aright|>epsilon_1$ for $L_1$

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Part 3

Then we use the condition from Part 1 and set $epsilon=frac{epsilon_1}{2}$

Therefore we get $delta>$ that $foralllambda<delta$ we have $left|sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i+sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Iright|<frac{epsilon_1}{2}$

According to Part 2 there is a partition and pick of points that makes $forall ARightarrowleft|sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Aright|>epsilon_1$

We make $L_1$ follow the above partition and pick of points, $L_2$ can be any partition and pick of points as long as $lambda<delta$

Thus we derive:

begin{aligned} because&frac{epsilon_1}{2}>left|sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i+sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Iright|\ =&left|sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-A+sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-(I-A)right|\ geq&left|left|sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Aright|-left|sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-(I-A)right|right|\ therefore&-frac{epsilon_1}{2}<left|sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-(I-A)right|-left|sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Aright|<frac{epsilon_1}{2}\ because&left|sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Aright|>epsilon_1\ therefore&left|sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-(I-A)right|>frac{epsilon_1}{2} end{aligned}

Note that the above condition only used the specific partition and pick of points of $L_1$. For $L_2$, despite of the partition and pick of points the result holds. Also dispute the value of $A$, the result holds.

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Part 4

Using condition 1 of Part 1, we define $left|vec{F}(vec{x})right|leq M$ thus we have $left|vec{F}(vec{x})cdotDeltavec{x}right|leq mleft|vec{F}(vec{x})right|left|Deltavec{x}right|leq nMleft|Deltavec{x}right|$

While $n$ is the dimension of $vec{F}(vec{x})$

Therefore, we have $left|sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_iright|leqsum_{i=k}^{m}left|vec{F}(vec{xi}_i)cdotDeltavec{x}_iright|<nMl_2$ where $l_2$ is the length of $L_2$ since $L$ is rectifiable.

We construct a partition of $P={vec{x}_k,...,vec{x}_{m+1}}$ separate $L_2$ equally on length into $m-k$ parts and pick of points $vec{xi}_i$ are the middle points of arc $vec{x}_ivec{x}_{i+1}$ on length.

Then donate $S_m=sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i,m>k$, with the previous proof we know that S_m is limited series.

Using Bolzano-Weierstrass theorem that there must be a sub-series that converges.

Donate $S_{m_u}$ that have $lim_{utoinfty}S_{m_u}=B$

This means that $forallepsilon>0,exists U,forall u>URightarrowleft|sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Bright|<epsilon$

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Part 5

From Part 4 we set $epsilon=frac{epsilon_1}{2}$ then we exists $U_1$

we set $u=maxleft(U_1,left[frac{l}{delta}right]+1right)$ then obviously we have $left|sum_{i=k}^{n}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Bright|<frac{epsilon_1}{2}$

From Part 3 we know that for any partition, pick of points, and $A$,

as long as $lambda=ldiv uleq ldiv left(left[frac{l}{delta}right]+1right)<delta$, for $L_2$ we have $left|sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-(I-A)right|>frac{epsilon_1}{2}$

We pick $A=I-B$ then have $lambda<delta$, for $L_2$ we have $left|sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i-Bright|>frac{epsilon_1}{2}$

Which result an contradiction.

Thus $lim_{lambdato0}sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i,lim_{lambdato0}sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_i$ both converge despite of partition and pick of points.

It is also easy to prove that they would converge into the same value by for any partition and pick of points $lim_{lambdato0}left{sum_{i=1}^{k-1}vec{F}(vec{xi}_i)cdotDeltavec{x}_i+sum_{i=k}^{m}vec{F}(vec{xi}_i)cdotDeltavec{x}_iright}$ always converge to $I$.

Thus it persuaded me that the existence are proved.