### CSIR JUNE 2011 PART C QUESTION 67 SOLUTION (If $f_n(x) \to 0$ (a.e) then $\int_a^b f_n(x) dx \to 0$)

Let $f_n$ be a sequence of integrable functions defined on an interval $[a,b]$. Then,
1) If $f_n(x) \to 0$ (a.e) then $\int_a^b f_n(x) dx \to 0$,
2) $\int_a^b f_n(x) \to 0$ then $f_n(x) \to 0$,
3) If $f_n(x) \to 0$ and each $f_n$ is a bounded function then $\int_a^b f_n(x) \to 0$,
4) If $f_n(x)\to 0$ and $f_n$'s are uniformly bounded then $\int_a^b f_n(x) dx \to 0$.
Solution: option 1: (False) Consider the sequence of functions on $[0,1]$ defined by $$f_n(x) = \begin{cases}n \text{ if } x \in [0,\frac{1}{n}] \\ 0 \text { if } x \in (\frac{1}{n}, 1] \end{cases}.$$
Claim: $f_n(x) \to 0$. Let $x \in (0,1]$ then there exists $n_0 \in \Bbb N$ such that $\frac{1}{n_0} < x$ and in particular $x \in (\frac{1}{n_0},1]$ (by using the archimedian property). For $n > n_0$ we have $\frac{1}{n}<\frac{1}{n+0}$ and so $x \in (\frac{1}{n},1]$. Hence $f_n(x) = 0$. This shows that the sequence $f_1(x),f_2(x),\dots$ is eventually zero. Hence $f_n(x) \to 0$
Now $\int_0^1 f_n(x) dx = \int_0^{\frac{1}{n}}n dx = n \int_0^{\frac{1}{n}}dx = 1$ for all $n$. Hence $\int_0^1 f_n(x)dx \to 1 \ne 0$.
option 2: (False) Define $f_n(x) = \begin{cases} \frac{1}{2} \text{ if } x = \frac{1}{2} \\ 0 \text{ otherwise} \end{cases}$ , then $f_n$ converge to the function $$f(x) = \begin{cases} \frac{1}{2} \text{ if } x = \frac{1}{2} \\ 0 \text{ otherwise }\end{cases}.$$
Clearly $\int_0^1 f_n(x)dx = 0$ for all $n$ and hence $\int_0^1f_n(x) \to 0$. But $f(x) \not\equiv 0$.
option 3: (False) Each of the function $f_n$ given in option (1) are bounded. So the same example works for this option also.
option 4: (True) (Because all the other options are false, but we will prove the result which is the correct way to do Mathematics)
Claim: Suppose $f_n$s are uniformly bounded, then there exists $k$ such that $|f_n(x)| \le k$ for all $n \in \Bbb N$ and $x \in [a,b]$. Let $g(x)$ be the constant function $k$ defined on $[a,b]$ then we have $|f_n(x)| \le g(x)$. Now, by the dominated convergence theorem, we can interschange the limit and the integral. Hence, we have $$\lim_{n \to \infty} \int_{a}^b f_n(x)dx = \int_a^b \lim_{n \to \infty}f_n(x) dx = \int_a^b 0 dx = 0.$$
### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft...