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I wish to show whether or not the sequences $f_n(x)=\chi_{[n,n+1]}$ is tight

By definition, Let $(X,m,\mu)$be a measure space, the sequence ${f_n}$ is called tight over $X$ if $\forall \epsilon>0$ $ \exists X_0 \subset X $,with $\mu (X_0)<\infty$ , such that $\forall n$ $\int_{X-X_0}|f_n|d\mu <\epsilon$.

Since for each $n$ the measure of the set $[n,n+1]$ $\mu([n,n+1])=1$, is finite in the first place, I chose $X_0=(n,n+1)$ so that $$\int_{X-x_0}|f_n|d\mu=\int_{[n,n+1]-(n,n+1)}|f_n|=\int_{\{n\}_{n=1}}|f_n|d\mu$$ And so, since the measure of the set $\{n\}_{n=1}$ is $0$, I will get $\int_{\{n\}_{n=1}}|f_n|d\mu=0$. and conclude that $f_n(x)=\chi_{[n,n+1]}$ is tight. I that right?

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$X_0$ must be independent of $n$, which is not the case in your example. I assume in the following that $\mu$ is the Lebesgue measure on the real line $\mathbb{R}$.

Let $X_0 \subset X$ such that $\mu(X_0)$ is finite.

Given any $\varepsilon > 0$, there exists $N_\varepsilon \in \mathbb{N}$ such that for all $n > N_\varepsilon$, one has $\mu( X_0 \cap [n,n+1]) < \varepsilon$.

Indeed, one has \begin{align} \mu(X_0) \geq \sum_{n\in \mathbb{N}} \mu(X_0 \cap [n, n+1])\end{align} Therefore $\mu(X_0 \cap [n, n+1]) \longrightarrow_n 0$ because $X_0$ is of finite measure and we're summing over positive numbers, so that the terms over which we sum must goes to zero positively. Therefore, for all $n > N_\varepsilon$ \begin{align}\mu(X\setminus X_0 \cap [n,n+1]) > 1 - \epsilon \end{align} It follows $(f_n)_n$ is not tight, as for every $0<\varepsilon<\frac{1}{2}$, given any $X_0 \subset X$ of finite measure, and for any $n$ big enough, we have: \begin{align} \int\limits_{X\setminus X_0} \chi_{[n, n +1 ]} d\mu = \mu(X\setminus X_0 \cap [n,n+1]) > 1 - \epsilon > \epsilon \end{align}

Intuitively, $\mu(X_0 \cap [n,n+1])$ must goes to $0$, otherwise we would have infinitely many disjoint subsets (i.e. the $X_0\cap[n,n+1]$) of $X_0$ of measure $>\varepsilon$, thus $X_0$ couldn't be of finite measure.

Beware, $\mu(X_0 \cap [n,n+1])$ needn't be exactly zero: consider $X_0 = \bigcup_{k=0}^{\infty} [k, k + \frac{1}{k^2}]$

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    Thank you. Even though I dont understand entirely your proof, By mentioning that $X_0$ does not depend on $n$ is very helpful. So, why did you not pick $N_{\epsilon }$ such that $x_0 \cap [n,n+1]=\phi$. so that $\mu (X_0\cap[n,n+1]=0)$Would that make it easier? if not can you explain intuitively why you made that choice? – 2017-02-14
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    I've made things a little bit clearer. As explained below, we do not necessarily have such a $N_{\varepsilon}$ – 2017-02-14
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    Okay, I am closer to understanding. thanks for the clarification. Using the example you gave for $X_0=\cup_{k=1}^{\infty}[n,n+\frac{1}{n^2}]$, Is it not enough to write $\int_{X\backslash X_0}f_n d\mu=\mu(X \backslash X_0)>1-\epsilon$? instead of $\int_{X\backslash X_0}f_n d\mu=\mu(X \backslash X_0\cup[n,n+1])>1-\epsilon$?Because f_n is only supported on $[n,n+1]$ and $x_0 \subset X$. so I feel it's redundant to write $X_0\cup[n,n+1]$ – 2017-02-14
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    sorry, I meant instead of $\mu(X\backslash X_0\cap[n,n+1])>1- \epsilon$ – 2017-02-14
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    This is false $\int_{X\backslash X_0}f_n d\mu \neq \mu(X \backslash X_0)$! Remember $\int_{A} g d\mu = \int_{\mathbb{R}} \chi_A \cdot g d \mu$ if $A \subset \mathbb{R}$ measurable and $g$ $ \mu$-measurable. Hence, $\int_{X\backslash X_0}\chi_{[n,n+1]} d\mu = \int_{X} \chi_{[n,n+1]} \cdot \chi_{X \backslash X_0} d\mu = \int_{X} \chi_{[n,n+1] \bigcap X \backslash X_0} d\mu = \mu(X \backslash X_0 \bigcap [n,n+1])$ – 2017-02-14
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    Since $X_0$ does not depend on $n$ can we write your example as $X_0=\cup_{k=0}^{\infty}[k,k+\frac{1}{k^2}]$ where $k\in N$. so that it looks fixed. – 2017-02-16
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    Sorry, I didn't realize I put $n$ instead of the $k's$. My mistake. – 2017-02-16
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Roughly speaking, a sequence of functions is tight if you can find a set $X_0$ with finite measure that contains 'nearly all' of the mass of every function.

We prove your sequence of functions is not tight using a contraposition argument. Suppose $X_0$ is a set such that $\int_{X-X_0}|f_n|d\mu <\epsilon$ for every $n$. Then, since each $f_n$ is nonnegative and integrates to $1$, we have $$\int_{X_0}f_n\,d\mu >1-\epsilon\qquad\text{for every $n$}.\tag{*}$$ Write $$ \mu(X_0)=\int_{X_0}1\,d\mu\stackrel{(1)}\ge\int_{X_0}\sum f_n\,d\mu\stackrel{(2)}=\sum\int_{X_0}f_n\,d\mu;\tag{**} $$ in (1) we use the fact that the $f_n$'s are indicators of intervals that are (almost) disjoint; in (2) we can interchange summation and integration because everything is non-negative. But the RHS of (**) will be arbitrarily large, in view of (*). Conclude that $X_0$ cannot have finite measure.

The intuition behind tightness is that 'no mass can escape' as $n\to\infty$. In your example it's clear that mass is indeed escaping, because the $f_n$'s are essentially unit masses marching off to the right.

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    I was thinking, Why do we write $\mu(X_0)=\int_ {X_0}1d\mu......$ as in $(**)$. Is this by definition or by construction – 2017-02-16
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    Also in $(1)$, I believe given the indicator functions, $f_n(x) \leq \sum f_n $ and so by Monotonicity of the integral $\int_{X_0}f_n=\int_{X_0}1\leq \int_{X_0}\sum f_n $ . Why is the inequality the other way? – 2017-02-16
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    @J.Kyei By definition, the integral of the simple function $\chi_A$ is $\mu(A)$. As for (1), $\sum f_n=\sum \chi_{[n,n+1)}=\chi_{[1,\infty)}$ and this is less than or equal to the constant function $1$. (For simplicity I'm using non-overlapping intervals, as the difference is a set of measure zero). How you get $\int_{X_0} f_n=\int_{X_0}1$? This is not true -- each $f_n$ is zero except on an interval of length 1. – 2017-02-16