I'm looking at an example here: https://www.math.ucdavis.edu/~romik/teaching-pages/mat235a-2013/discussion8.pdf (2nd example below Lemma 5)

We have that $X_n$ converges in probability to 0. I'm looking at the bit where we show want to $X_n/(1+X_n)$ converges in $L^1$ to 0.

I do not understand how the author has bounded the expectation above by what is shown. I have tried to use Markov's inequality, but I can only bound it below.

  • $\begingroup$ You might be missing the fact that $|X_n|/(1+|X_n|) < 1$. This is important to getting the upper bound. $\endgroup$
    – fblundun
    Nov 25, 2020 at 11:35
  • $\begingroup$ @fblundun so it is bounded above by 1, which is also $P(|X_n| > \epsilon) + P(|X_n| \leq \epsilon) $, but where does the term before $P(|X_n| \leq \epsilon)$ come from? $\endgroup$
    – Crowd
    Nov 25, 2020 at 11:48

1 Answer 1


Let $Y_n = \frac{|X_n|}{1+|X_n|}$. Then $Y_n$ is bounded above by 1.

For $\epsilon\gt0$,

$$ \begin{align}\mathbb{E}(Y_n)&\leq \left(\mathbb{P}(|X_n|\lt\epsilon)\sup_{|X_n|\lt\epsilon}Y_n\right) + \left(\mathbb{P}(|X_n|\ge\epsilon)\sup_{|X_n|\ge\epsilon}Y_n\right)\\& \leq\left(\sup_{|X_n|\lt\epsilon}Y_n\right)+\mathbb{P}(|X_n|\geq\epsilon)\\ &= \frac{\epsilon}{1+\epsilon} + \mathbb{P}(|X_n|\geq\epsilon)\\ &\lt\epsilon + \mathbb{P}(|X_n|\geq\epsilon) \end{align} $$

The second term converges to 0 so this can be made arbitrarily small.


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