Let us consider the following scenario: we have a sequence $(X_n)_{n\geq 1}$ of independent random variables, where for every integer $n\geq 1$, $X_n$ is exponentially distributed with parameter $\lambda_n > 0$. Let us define the random variable $m_n = \min \{X_1, \ldots, X_n\}$ for any integer $n\geq 1$. We need to show that if $\sum\limits_{n=1}^{\infty}\lambda_n = \lambda <\infty$, then, with probability 1, the minimum record gets broken only finitely many times, and if that same $\lambda$ is $\infty$, then, with probability 1, the record gets broken infinitely many times.
My attempt:
It is known that $m_n$ is also exponential with parameter $\lambda_1 + \ldots + \lambda_n$ (for any positive integer $n$). Thus, one can easily see that $m_n \xrightarrow{d}\text{EXP}(\lambda)$, if $\lambda<\infty$, and $m_n \xrightarrow{d} 0$ if $\lambda=\infty$.
Now, let $N_n$ be the number of times the minimum record is broken up to step $n$, i.e. $$N_n = \sum_{i=2}^n \mathbb{I}_{\{m_i = X_i\}}=\sum_{i=2}^n\mathbb{I}_{\{X_i < m_{i-1}\}},$$ where $\mathbb{I}_E$ is the indicator of an event $E$.
Since $m_i\sim \text{EXP}\left(\sum\limits_{k=1}^i \lambda_k\right)$, we know that $\mathbb{P}(X_i < m_{i-1}) = \frac{\lambda_i}{\lambda_1 + \ldots + \lambda_i}$, but I don't seem to think these events are independent, and even if they are, I am not sure how to proceed with finding the probability of $N_n$ being bounded when $n\to\infty$.
Any help would be appreciated!
I think your work is already in the right direction.
By the (first) Borel–Cantelli lemma (which doesn’t require independence), the probability of infinitely many records being set is $0$ if the sum of the probabilities of those events is finite. And indeed
\begin{eqnarray*} \sum_i\mathbb P(X_i\lt m_{i-1}) &=& \sum_i\frac{\lambda_i}{\lambda_1+\cdots+\lambda_i} \\ &\le& \sum_i\frac{\lambda_i}{\lambda_1} \\ &=& \frac\lambda{\lambda_1} \\ &\lt& \infty\;. \end{eqnarray*} For the other result, we can’t use the second Borel–Cantelli lemma (at least I don’t see how to) because that does require independence. Instead, consider the events $B_n$ that the $X_i$ never go below $\frac1n$. Their probability is $0$:
\begin{eqnarray*} \mathbb P(B_n) &=& \prod_i\mathbb P\left(X_i\ge\frac1n\right) \\ &=& \prod_i\exp\left(-\lambda_i\cdot\frac1n\right) \\ &=& \exp\left(-\sum_i\lambda_i\cdot\frac1n\right) \\ &=&0\;. \end{eqnarray*}
Thus, the probability of the union of these countably many events is also $0$. But the event that there are finitely many records is contained in this union (since the last record is a lower bound on the $X_i$). Thus it also has probability $0$, so the probability for the record to be broken infinitely often is $1$.