Decreasing the weight of one edge of minimum spanning tree, prove the MST is unchanged
There is no necessarily a unique MST for a given G. So reducing an edge might trigger an MST finding algorithm to find a alternative MST. The question is if given MST remains an MST after the change.

Let's make some definitions:


$e$ is the edge we reduce its weight
$w(e)$ is the weight of the edge e
$w(T)$ is the weight of tree T (sum of all edges)


I will prove it by contradiction
Let's suppose that after decreasing the weight of of edge $e$ from $w(e)$ to $w'(e)$ T is no more a minimal spanning tree, and $T'$ is a minimal spanning tree.


First we consider the case where $e \in T'$


After the decrease of w(e)-> w'(e) the new graph is $G'$.
In $G'$ we have:

$w(T') &lt; w(T)$ by assumption

$w'(e) + w(T' - {e}) &lt; w'(e) + w(T-{e})$

$w(T'-{e}) &lt; w(T-{e})$ and that is true both in G and G', since $\{T'-e\}$ and $\{T-{e}\}$ are same in both graphs

Then subsequently in $G$ we have
$w(T'-{e}) &lt; w(T-{e})$

$w(e) + w(T'-{e}) &lt; w(e)+ w(T-{e})$

$w(T') &lt; w(T)$ which means that in G T is not a minimal spanning tree which is not true.


Now we consider the case where $e \notin T'$


Then in G' $w(T')&lt; w(T)$ and by the same sequence 

$w(T'-{e}) &lt;  w(T-e)$ in both $G$ and $G'$

In $G$ we have consequently
$w(T'-{e}) + w(e) &lt; w(T-{e}) + w(e)$ and if $T'' \equiv \{T'+e\}$ then

$w(T'') &lt; w(T)$ which again contradicts the truth that T is MST in G.