Prove that a function is continuous under 1 dimensional Lebesgue measure

Question

Consider the measure space $(\Bbb R, \Sigma, \lambda )$, where is a measure on $\Bbb R$ and is the associated $\sigma$-algebra; assume that contains $B(\Bbb R)$ (the Borel $\sigma$-algebra).

Let $u : \Bbb R \to \overline{\Bbb R}$ be a $\Sigma$-measurable and $\lambda$-summable function; consider the function

$v : (0,1) \to \Bbb R$ defined by $x \to v(x) := \int_{[0,x]}u d\lambda$ (for $x > 0$)

Assume that $\lambda$ is the Lebesgue 1-dimensional measure. Prove that the function $v$ is continuous on $(0,1)$.

So, the question asks that I need to show that $x_j \to x \implies v(x_j) \to v(x)$.

We know that $\int_{[a,b]} u d\lambda = \int_{\Bbb R} u \Bbb 1[a,b] d\lambda$

I have to use some form of convergence theorem here and I assume it would have to be Monotone convergence as then I can split the limits into 2 cases, from left and from right $x_j \to x^+$ and $x_j \to x^-$.

I am just not sure how to formalise this or if that is the right approach to begin with.

Answer 1

The sequence $\left\{u\cdot1_{(0,x_j]}\right\}_{j\in\Bbb N}$ is dominated by $\lvert u\rvert\in L^1(0,1)$ and it converges pointwise to $u\cdot 1_{(0,x)}$. So you can use dominated convergence on this sequence (and any other sequence, since the pointwise limit won't change).

Answer 2

It is enough to consider the case $u \geq 0$ because $v(x)=\int_{[0,x]} u^{+}d\lambda -\int_{[0,x]} u^{-}d\lambda$. Suppose $x_j $ increases to $x$. Then $I_{(x_j,x]}$ decreases to $I_{\{x\}}$. Hence $1-I_{(x_j,x]}$ increases to $1-I_{\{x\}}$. Apply Monotone Convergence theorem to this sequence and subtract $\int u d\lambda$ from both sides. The case $x_j $ decreasing to $x$ is similar. Note: the fact that $\lambda (\{x\})=0$ is important in this proof.